Trigonometric Equations

Study Sheet

Trigonometric Equations

Basic equations, factoring, quadratic form, identities, multiple angles, and inverses

Solving Basic Trigonometric Equations

A trigonometric equation is true only for certain values of the variable. Because sine, cosine, and tangent are periodic, most equations have infinitely many solutions. You will usually be asked for one of two things:

  • All solutions — write a general formula using +2πn+2\pi n (for sin⁡\sin/cos⁡\cos) or +πn+\pi n (for tan⁡\tan), where nn is any integer.
  • Solutions on an interval, usually [0,2π)[0,2\pi) — list only the values inside that interval.

Solving a trig equation means finding every angle in a given interval that satisfies it. Isolate the function (or factor so each factor is a single function), find the reference angle from the exact value, place it in the quadrants where the sign is right, and add the period for the general solution. Equations in 2x2x are solved for 2x2x over a doubled interval first.

Concept
The idea

To solve sin⁡x=k\sin x = k (or cos⁡x=k\cos x = k, tan⁡x=k\tan x = k):

  • Find the reference angle from the special-angle table (or arcsin⁡\arcsin, arccos⁡\arccos, arctan⁡\arctan).
  • Use the sign of kk to decide which quadrants the solutions live in.
  • Add the period to get all solutions: +2πn+2\pi n for sine/cosine, +πn+\pi n for tangent.

Useful periods: sin⁡x=0⇒x=πn\sin x = 0 \Rightarrow x=\pi n;   cos⁡x=0⇒x=π2+πn\cos x = 0 \Rightarrow x=\tfrac{\pi}{2}+\pi n;   tan⁡x=0⇒x=πn\tan x = 0 \Rightarrow x=\pi n.

Example
Worked example: all solutions and solutions on [0,2π)[0,2\pi)
xy−11π/32π/3π4π/35π/3y = −1/2
Two crossings per period; every other solution is one of these plus a multiple of 2π2\pi.

Solve 2cos⁡x+1=02\cos x + 1 = 0.

2cos⁡x+1=0cos⁡x=−12\begin{aligned} 2\cos x + 1 &= 0 \\ \cos x &= -\tfrac{1}{2} \end{aligned}

Reference angle: π3\dfrac{\pi}{3}. Cosine is negative in Quadrants II and III, so on [0,2π)[0,2\pi)

x=2π3,x=4π3.x = \frac{2\pi}{3}, \qquad x = \frac{4\pi}{3}.

All solutions:   x=2π3+2πnorx=4π3+2πn,n∈Z.x = \dfrac{2\pi}{3} + 2\pi n \quad\text{or}\quad x = \dfrac{4\pi}{3} + 2\pi n, \quad n\in\mathbb{Z}.

Side note
Where it goes wrong

Dividing both sides by sin⁡x\sin x and losing the solutions where sin⁡x=0\sin x = 0. Factor instead: sin⁡xcos⁡x=sin⁡x\sin x\cos x = \sin x becomes sin⁡x(cos⁡x−1)=0\sin x(\cos x - 1) = 0.

The picture shows why sin⁡x=12\sin x = \tfrac12 has two solutions on [0,2π)[0,2\pi): the curve y=sin⁡xy=\sin x crosses the line y=12y=\tfrac12 twice.

Tip

Don't forget the period. A common mistake is giving only one solution. For sine/cosine always attach +2πn+2\pi n; for tangent attach +πn+\pi n. To restrict to [0,2π)[0,2\pi), keep adding the period until you leave the interval.

Try it
Quick check

Solve 2sin⁡x+3=02\sin x+\sqrt3=0 on [0,2π)[0,2\pi), then give all solutions.

Show the answer

Answer: x=4π3, 5π3x=\dfrac{4\pi}{3},\ \dfrac{5\pi}{3}; all solutions x=4π3+2πnx=\dfrac{4\pi}{3}+2\pi n or x=5π3+2πnx=\dfrac{5\pi}{3}+2\pi n. Isolate: sin⁡x=−32\sin x=-\dfrac{\sqrt3}{2}. The reference angle is π3\dfrac{\pi}{3}, and sine is negative in Quadrants III and IV: π+π3\pi+\dfrac{\pi}{3} and 2π−π32\pi-\dfrac{\pi}{3}.

Equations Solved by Factoring

Concept
The idea

Move every term to one side so the equation equals 00, factor, then set each factor equal to 00 (the Zero-Product Property). Solve each simple equation separately and combine the answers.

A product is zero exactly when one of its factors is zero, so factoring splits one hard equation into several basic ones without losing anything. Dividing by a factor, by contrast, silently assumes that factor is not zero, and the solutions that make it zero disappear.

Example
Worked example: a full factoring solve

Solve 2sin⁡xcos⁡x=cos⁡x2\sin x\cos x = \cos x on [0,2π)[0,2\pi).

2sin⁡xcos⁡x−cos⁡x=0(get one side =0)cos⁡x (2sin⁡x−1)=0(factor out cos⁡x)\begin{aligned} 2\sin x\cos x - \cos x &= 0 &&\text{(get one side $=0$)}\\ \cos x\,(2\sin x - 1) &= 0 &&\text{(factor out $\cos x$)} \end{aligned}

Set each factor to zero:

cos⁡x=0  ⇒  x=π2, 3π2;2sin⁡x−1=0  ⇒  sin⁡x=12  ⇒  x=π6, 5π6.\cos x = 0 \;\Rightarrow\; x=\frac{\pi}{2},\ \frac{3\pi}{2}; \qquad 2\sin x - 1 = 0 \;\Rightarrow\; \sin x=\tfrac12 \;\Rightarrow\; x=\frac{\pi}{6},\ \frac{5\pi}{6}.

Solutions:   x=π6, π2, 5π6, 3π2.x = \dfrac{\pi}{6},\ \dfrac{\pi}{2},\ \dfrac{5\pi}{6},\ \dfrac{3\pi}{2}.

Tip

Never divide both sides by a trig factor. Dividing 2sin⁡xcos⁡x=cos⁡x2\sin x\cos x=\cos x by cos⁡x\cos x throws away the solutions where cos⁡x=0\cos x=0. Always factor instead.

Try it
Quick check

Solve sin⁡xtan⁡x=sin⁡x\sin x\tan x=\sin x on [0,2π)[0,2\pi).

Show the answer

Answer: x=0, π4, π, 5π4x=0,\ \dfrac{\pi}{4},\ \pi,\ \dfrac{5\pi}{4}. Do not divide by sin⁡x\sin x. Factor: sin⁡x(tan⁡x−1)=0\sin x(\tan x-1)=0. Then sin⁡x=0\sin x=0 gives 0,π0,\pi, and tan⁡x=1\tan x=1 gives π4,5π4\dfrac{\pi}{4},\dfrac{5\pi}{4}. All four keep tan⁡x\tan x defined.

Quadratic-Form Equations (Substitution)

Concept
The idea

If the equation looks like a quadratic in one trig function — for example 2sin⁡2x−sin⁡x−1=02\sin^2 x - \sin x - 1 = 0 — substitute uu for that function, solve the quadratic (factor or quadratic formula), then back-substitute. Discard any value with ∣u∣>1|u|>1 for sine or cosine (impossible).

Reminder — The quadratic formula:  x=−b±b2−4ac2a(a≠0)x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\qquad (a\neq 0)
Example
Worked example: substitution

Solve 2sin⁡2x−sin⁡x−1=02\sin^2 x - \sin x - 1 = 0 on [0,2π)[0,2\pi). Let u=sin⁡xu=\sin x:

2u2−u−1=0(2u+1)(u−1)=0u=−12oru=1.\begin{aligned} 2u^2 - u - 1 &= 0 \\ (2u+1)(u-1) &= 0 \\ u = -\tfrac12 \quad &\text{or}\quad u = 1. \end{aligned}

Back-substitute:

sin⁡x=1  ⇒  x=π2;sin⁡x=−12  ⇒  x=7π6, 11π6.\sin x = 1 \;\Rightarrow\; x=\frac{\pi}{2}; \qquad \sin x = -\tfrac12 \;\Rightarrow\; x=\frac{7\pi}{6},\ \frac{11\pi}{6}.

Solutions:   x=π2, 7π6, 11π6.x = \dfrac{\pi}{2},\ \dfrac{7\pi}{6},\ \dfrac{11\pi}{6}.

Tip

Check the range. A factor like cos⁡x=2\cos x = 2 has no solution, since cosine never exceeds 11. Simply drop that factor and keep the others.

Try it
Quick check

Solve 2cos⁡2x−3cos⁡x+1=02\cos^2x-3\cos x+1=0 on [0,2π)[0,2\pi).

Show the answer

Answer: x=0, π3, 5π3x=0,\ \dfrac{\pi}{3},\ \dfrac{5\pi}{3}. With u=cos⁡xu=\cos x: 2u2−3u+1=(2u−1)(u−1)=02u^2-3u+1=(2u-1)(u-1)=0, so u=12u=\tfrac12 or u=1u=1. cos⁡x=12\cos x=\tfrac12 gives π3,5π3\dfrac{\pi}{3},\dfrac{5\pi}{3}; cos⁡x=1\cos x=1 gives only x=0x=0 (a boundary value, one solution).

Equations Requiring an Identity First

Concept
The idea

When an equation mixes different functions or has a squared term, rewrite it in terms of a single trig function using an identity, then factor. Most common:

sin⁡2x=1−cos⁡2x,cos⁡2x=1−sin⁡2x,cos⁡2x=1−2sin⁡2x=2cos⁡2x−1.\sin^2 x = 1-\cos^2 x, \qquad \cos^2 x = 1-\sin^2 x, \qquad \cos 2x = 1-2\sin^2 x = 2\cos^2 x - 1.
Example
Worked example: Pythagorean identity first

Solve 2cos⁡2x+sin⁡x−1=02\cos^2 x + \sin x - 1 = 0 on [0,2π)[0,2\pi). Replace cos⁡2x=1−sin⁡2x\cos^2 x = 1-\sin^2 x:

2(1−sin⁡2x)+sin⁡x−1=0−2sin⁡2x+sin⁡x+1=02sin⁡2x−sin⁡x−1=0(2sin⁡x+1)(sin⁡x−1)=0.\begin{aligned} 2(1-\sin^2 x) + \sin x - 1 &= 0 \\ -2\sin^2 x + \sin x + 1 &= 0 \\ 2\sin^2 x - \sin x - 1 &= 0 \\ (2\sin x + 1)(\sin x - 1) &= 0. \end{aligned}

So sin⁡x=−12⇒x=7π6, 11π6\sin x = -\tfrac12 \Rightarrow x=\dfrac{7\pi}{6},\ \dfrac{11\pi}{6} and sin⁡x=1⇒x=π2\sin x = 1 \Rightarrow x=\dfrac{\pi}{2}. Solutions:   x=π2, 7π6, 11π6.x = \dfrac{\pi}{2},\ \dfrac{7\pi}{6},\ \dfrac{11\pi}{6}.

Reminder — The Pythagorean identity:  sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1
Tip

One function at a time. You cannot factor an equation that still contains both sin⁡\sin and cos⁡\cos squared terms. Convert everything to a single function before you try to factor.

Try it
Quick check

Solve cos⁡2x=cos⁡x\cos 2x=\cos x on [0,2π)[0,2\pi).

Show the answer

Answer: x=0, 2π3, 4π3x=0,\ \dfrac{2\pi}{3},\ \dfrac{4\pi}{3}. Use the cosine-only form cos⁡2x=2cos⁡2x−1\cos2x=2\cos^2x-1: 2cos⁡2x−cos⁡x−1=02\cos^2x-\cos x-1=0, which factors as (2cos⁡x+1)(cos⁡x−1)=0(2\cos x+1)(\cos x-1)=0. So cos⁡x=−12\cos x=-\tfrac12 (x=2π3,4π3x=\tfrac{2\pi}{3},\tfrac{4\pi}{3}) or cos⁡x=1\cos x=1 (x=0x=0).

Multiple-Angle Equations

Concept
The idea

For an equation in 2x2x, 3x3x, or x2\tfrac{x}{2}, first solve for the whole angle, then divide. If x∈[0,2π)x\in[0,2\pi) then:

2x∈[0,4π),3x∈[0,6π),x2∈[0,π).2x \in [0,4\pi), \qquad 3x \in [0,6\pi), \qquad \tfrac{x}{2}\in[0,\pi).

List all angle solutions across the expanded interval, then divide each by the multiple.

Example
Worked example: list all multiple-angle solutions

Solve 2cos⁡(2x)=12\cos(2x) = 1 on [0,2π)[0,2\pi).

cos⁡(2x)=12.\begin{aligned} \cos(2x) &= \tfrac12. \end{aligned}

Let θ=2x\theta = 2x. Since x∈[0,2π)x\in[0,2\pi), we need θ∈[0,4π)\theta\in[0,4\pi). Cosine equals 12\tfrac12 at reference angle π3\tfrac{\pi}{3} in Quadrants I and IV, over two full turns:

θ=π3, 5π3, 7π3, 11π3.\theta = \frac{\pi}{3},\ \frac{5\pi}{3},\ \frac{7\pi}{3},\ \frac{11\pi}{3}.

Now x=θ2x=\dfrac{\theta}{2}:

x=π6, 5π6, 7π6, 11π6.x = \frac{\pi}{6},\ \frac{5\pi}{6},\ \frac{7\pi}{6},\ \frac{11\pi}{6}.

General solution:   x=π6+πnorx=5π6+πn.x = \dfrac{\pi}{6} + \pi n \quad\text{or}\quad x = \dfrac{5\pi}{6} + \pi n.

Tip

Expand the interval first. Because 2x2x sweeps around twice as fast, you get twice as many solutions. Solve for the angle over [0,4π)[0,4\pi) (or [0,6π)[0,6\pi) for 3x3x) before dividing — otherwise you will lose half the answers.

Try it
Quick check

Solve tan⁡3x=1\tan 3x=1 on [0,π)[0,\pi).

Show the answer

Answer: x=π12, 5π12, 3π4x=\dfrac{\pi}{12},\ \dfrac{5\pi}{12},\ \dfrac{3\pi}{4}. Let θ=3x∈[0,3π)\theta=3x\in[0,3\pi). Tangent is 11 at π4+πn\dfrac{\pi}{4}+\pi n: θ=π4, 5π4, 9π4\theta=\dfrac{\pi}{4},\ \dfrac{5\pi}{4},\ \dfrac{9\pi}{4} (the next, 13π4\tfrac{13\pi}{4}, is past 3π3\pi). Divide by 33.

Inverse Functions and Extraneous Solutions

Concept
The idea

Inverse functions: when the value is not a special angle, use arcsin⁡\arcsin, arccos⁡\arccos, or arctan⁡\arctan on a calculator. It returns only one angle; use quadrant symmetry and the period to find the rest.

Extraneous solutions: squaring both sides or multiplying by a trig factor can create false solutions. Always substitute each candidate back into the original equation and discard any that fail (or make a function undefined).

Example
Worked example: inverse function

Solve tan⁡x=3\tan x = 3 on [0,2π)[0,2\pi) (round to three decimals). The reference is arctan⁡3≈1.249\arctan 3 \approx 1.249. Tangent is positive in Quadrants I and III:

x≈1.249,x≈1.249+π≈4.391.x \approx 1.249, \qquad x \approx 1.249 + \pi \approx 4.391.
Example
Worked example: checking for extraneous solutions
xy−11π/2π3π/2y = 1
y=sin⁡x−cos⁡xy=\sin x-\cos x meets y=1y=1 only at π2\tfrac{\pi}{2} and π\pi; the rejected candidates (open dots) sit at y=−1y=-1.

Solve sin⁡x−cos⁡x=1\sin x - \cos x = 1 on [0,2π)[0,2\pi). Square both sides:

(sin⁡x−cos⁡x)2=12sin⁡2x−2sin⁡xcos⁡x+cos⁡2x=11−2sin⁡xcos⁡x=1sin⁡xcos⁡x=0.\begin{aligned} (\sin x - \cos x)^2 &= 1^2 \\ \sin^2 x - 2\sin x\cos x + \cos^2 x &= 1 \\ 1 - 2\sin x\cos x &= 1 \\ \sin x\cos x &= 0. \end{aligned}

Candidates: sin⁡x=0⇒x=0,π\sin x = 0 \Rightarrow x=0,\pi; cos⁡x=0⇒x=π2,3π2\cos x=0 \Rightarrow x=\tfrac{\pi}{2},\tfrac{3\pi}{2}. Check each in the original:

x=0:0−1=−1≠1(reject)x=π2:1−0=1✓x=π:0−(−1)=1✓x=3π2:−1−0=−1≠1(reject)\begin{array}{llll} x=0: & 0-1=-1 \ne 1 & \text{(reject)} & \\ x=\tfrac{\pi}{2}: & 1-0=1 \checkmark & x=\pi: & 0-(-1)=1 \checkmark \\ x=\tfrac{3\pi}{2}: & -1-0=-1 \ne 1 & \text{(reject)} & \end{array}

Solutions:   x=π2, π.x = \dfrac{\pi}{2},\ \pi.

Tip

Always verify after squaring or multiplying. These steps are not reversible — substitute every candidate into the original equation, and throw out any that do not check or that make tan⁡\tan, sec⁡\sec, etc. undefined.

Try it
Quick check

Solve 3sin⁡x=23\sin x=2 on [0,2π)[0,2\pi), to three decimal places.

Show the answer

Answer: x≈0.730x\approx 0.730 and x≈2.412x\approx 2.412. sin⁡x=23\sin x=\tfrac23. The calculator gives arcsin⁡23≈0.7297\arcsin\tfrac23\approx 0.7297 (Quadrant I). Sine is also positive in Quadrant II, at π−0.7297≈2.4119\pi-0.7297\approx 2.4119.

Going Deeper: Advanced Trig Equations

The equations below combine every earlier technique. The recurring theme is convert first, solve second: reshape the equation with an identity or an RR-form until it becomes one of the basic types you already know.

Concept
Sum-to-product and double-angle as an opening move

Some equations cannot be factored until you first merge or split terms. Two toolkits do this:

  • Sum-to-product turns a sum of two sines/cosines into a product, which then factors: @@BLOCK0@@
  • Double-angle rewrites sin⁡2x\sin 2x or cos⁡2x\cos 2x so a mixed equation collapses to a single function: @@BLOCK1@@

Rule of thumb: a sum of trig terms →\to sum-to-product; a lone sin⁡2x\sin 2x/cos⁡2x\cos 2x mixed with sin⁡x\sin x/cos⁡x\cos x →\to double-angle.

Example
Worked example: sum-to-product first

Solve cos⁡3x+cos⁡x=0\cos 3x + \cos x = 0 on [0,2π)[0,2\pi). With A=3xA=3x, B=xB=x, so A+B2=2x\tfrac{A+B}{2}=2x and A−B2=x\tfrac{A-B}{2}=x:

cos⁡3x+cos⁡x=02cos⁡2x cos⁡x=0(sum-to-product)cos⁡2x=0orcos⁡x=0.\begin{aligned} \cos 3x + \cos x &= 0 \\ 2\cos 2x\,\cos x &= 0 &&\text{(sum-to-product)}\\ \cos 2x = 0 \quad &\text{or}\quad \cos x = 0. \end{aligned}

For cos⁡x=0\cos x = 0: x=π2, 3π2x=\dfrac{\pi}{2},\ \dfrac{3\pi}{2}. For cos⁡2x=0\cos 2x = 0, let θ=2x∈[0,4π)\theta=2x\in[0,4\pi), so θ=π2,3π2,5π2,7π2\theta=\dfrac{\pi}{2},\dfrac{3\pi}{2},\dfrac{5\pi}{2},\dfrac{7\pi}{2} and hence x=π4,3π4,5π4,7π4x=\dfrac{\pi}{4},\dfrac{3\pi}{4},\dfrac{5\pi}{4},\dfrac{7\pi}{4}.

Solutions: x=π4, π2, 3π4, 5π4, 3π2, 7π4.\textbf{Solutions: } x = \frac{\pi}{4},\ \frac{\pi}{2},\ \frac{3\pi}{4},\ \frac{5\pi}{4},\ \frac{3\pi}{2},\ \frac{7\pi}{4}.
Concept
Counting solutions of a multiple-angle equation

For sin⁡(kx)=c\sin(kx)=c or cos⁡(kx)=c\cos(kx)=c with kk a positive integer and −1<c<1-1<c<1, the whole angle kxkx ranges over [0,2πk)[0,2\pi k) when x∈[0,2π)x\in[0,2\pi). Each full turn contributes two solutions, so there are exactly 2k2k solutions on [0,2π)[0,2\pi). (For tan⁡(kx)=c\tan(kx)=c there are also 2k2k solutions: tan⁡\tan takes each value once per half-turn, and kxkx covers 2k2k half-turns.) This lets you predict the count before solving, then check you found them all.

Example
Worked example: find every solution and count them
xy−11π/2π3π/2
Three cycles of sin⁡3x\sin 3x on [0,2π)[0,2\pi), two crossings each: six solutions.

Solve sin⁡(3x)=32\sin(3x)=\tfrac{\sqrt3}{2} on [0,2π)[0,2\pi). Here k=3k=3, so expect 2k=62k=6 solutions. Let θ=3x∈[0,6π)\theta=3x\in[0,6\pi). Sine equals 32\tfrac{\sqrt3}{2} at reference angle π3\tfrac{\pi}{3} in Quadrants I and II, i.e. θ=π3\theta=\tfrac{\pi}{3} and θ=2π3\theta=\tfrac{2\pi}{3}, repeated every 2π2\pi across three turns:

θ=π3, 2π3, 7π3, 8π3, 13π3, 14π3.\theta = \frac{\pi}{3},\ \frac{2\pi}{3},\ \frac{7\pi}{3},\ \frac{8\pi}{3},\ \frac{13\pi}{3},\ \frac{14\pi}{3}.

Divide by 33 (i.e. x=θ/3x=\theta/3):

Solutions: x=π9, 2π9, 7π9, 8π9, 13π9, 14π9.\textbf{Solutions: } x = \frac{\pi}{9},\ \frac{2\pi}{9},\ \frac{7\pi}{9},\ \frac{8\pi}{9},\ \frac{13\pi}{9},\ \frac{14\pi}{9}.

That is exactly 66 solutions, as predicted.

Concept
The RR-form: solving asin⁡x+bcos⁡x=ca\sin x + b\cos x = c

A sum of a sine and a cosine with the same angle can be rewritten as a single sinusoid:

asin⁡x+bcos⁡x=Rsin⁡(x+φ),R=a2+b2,tan⁡φ=ba.a\sin x + b\cos x = R\sin(x+\varphi), \qquad R=\sqrt{a^2+b^2}, \qquad \tan\varphi = \frac{b}{a}.

Choose φ\varphi in the quadrant matching the signs of aa (as the “cosine” part) and bb (as the “sine” part). The equation becomes Rsin⁡(x+φ)=cR\sin(x+\varphi)=c, i.e. sin⁡(x+φ)=cR\sin(x+\varphi)=\tfrac{c}{R}, a basic equation. Existence check: real solutions exist only when ∣c∣≤R=a2+b2|c|\le R=\sqrt{a^2+b^2}.

Example
Worked example: the RR-form
xy−2−11π/2π3π/2y = 1
sin⁡x+3cos⁡x\sin x+\sqrt3\cos x is the sinusoid 2sin⁡ ⁣(x+π3)2\sin\!\left(x+\tfrac{\pi}{3}\right); it meets y=1y=1 at π2\tfrac{\pi}{2} and 11π6\tfrac{11\pi}{6}.

Solve sin⁡x+3 cos⁡x=1\sin x + \sqrt3\,\cos x = 1 on [0,2π)[0,2\pi). Here a=1a=1, b=3b=\sqrt3, so

R=1+3=2,tan⁡φ=31⇒φ=π3.R=\sqrt{1+3}=2, \qquad \tan\varphi = \frac{\sqrt3}{1} \Rightarrow \varphi=\frac{\pi}{3}.

Both a>0a>0 and b>0b>0, so φ=π3\varphi=\tfrac{\pi}{3} (Quadrant I). The equation becomes

2sin⁡ ⁣(x+π3)=1  ⇒  sin⁡ ⁣(x+π3)=12.2\sin\!\Big(x+\frac{\pi}{3}\Big)=1 \;\Rightarrow\; \sin\!\Big(x+\frac{\pi}{3}\Big)=\frac12.

Let u=x+π3u=x+\tfrac{\pi}{3}. Since x∈[0,2π)x\in[0,2\pi), u∈[π3, 2π+π3)u\in\big[\tfrac{\pi}{3},\,2\pi+\tfrac{\pi}{3}\big). Sine is 12\tfrac12 at u=π6,5π6,13π6u=\tfrac{\pi}{6},\tfrac{5\pi}{6},\tfrac{13\pi}{6}; keep those inside the uu-interval: u=5π6u=\tfrac{5\pi}{6} and u=13π6u=\tfrac{13\pi}{6}. Then x=u−π3x=u-\tfrac{\pi}{3}:

Solutions: x=π2, 11π6.\textbf{Solutions: } x = \frac{\pi}{2},\ \frac{11\pi}{6}.
Example
Worked example: quadratic in sin⁡\sin with careful casework

Solve 2sin⁡2x−3sin⁡x+1=02\sin^2 x - 3\sin x + 1 = 0 on [0,2π)[0,2\pi). Let u=sin⁡xu=\sin x:

2u2−3u+1=0(2u−1)(u−1)=0u=12oru=1.\begin{aligned} 2u^2 - 3u + 1 &= 0 \\ (2u-1)(u-1) &= 0 \\ u = \tfrac12 \quad &\text{or}\quad u = 1. \end{aligned}

Now handle each case separately and note how many solutions each yields:

sin⁡x=12:two solutionsx=π6, 5π6,sin⁡x=1:one solution (a peak)x=π2.\begin{array}{lll} \sin x = \tfrac12: & \text{two solutions} & x=\dfrac{\pi}{6},\ \dfrac{5\pi}{6},\\[4pt] \sin x = 1: & \text{one solution (a peak)} & x=\dfrac{\pi}{2}. \end{array}

A boundary value like sin⁡x=1\sin x = 1 gives a single solution because the curve just touches the line, not crosses it.

Solutions: x=π6, π2, 5π6(three total).\textbf{Solutions: } x = \frac{\pi}{6},\ \frac{\pi}{2},\ \frac{5\pi}{6} \quad(\text{three total}).
Tip

Boundary cases count once. When a quadratic in sin⁡x\sin x or cos⁡x\cos x produces u=±1u=\pm1, that value gives one solution on [0,2π)[0,2\pi), not two, because sin⁡x=1\sin x=1, sin⁡x=−1\sin x=-1, cos⁡x=1\cos x=1, cos⁡x=−1\cos x=-1 each occur at a single point. Values with ∣u∣<1|u|<1 give two. Miscounting here is the most common error in “how many solutions” problems.

Concept
Parameter conditions: how many solutions does sin⁡x=k\sin x = k have?

Think of solving f(x)=kf(x)=k as intersecting the graph y=f(x)y=f(x) with the horizontal line y=ky=k. For sin⁡x=k\sin x = k on [0,2π)[0,2\pi):

∣k∣>1:no solution (line misses the curve),k=1 or k=−1:exactly one solution (line is tangent at a peak/valley),−1<k<1, k≠0:two solutions,k=0:two solutions on [0,2π) (x=0,π).\begin{array}{ll} |k|>1: & \text{no solution (line misses the curve)},\\ k=1 \text{ or } k=-1: & \text{exactly one solution (line is tangent at a peak/valley)},\\ -1<k<1,\ k\ne 0: & \text{two solutions},\\ k=0: & \text{two solutions on } [0,2\pi)\ (x=0,\pi). \end{array}

The same graph-versus-line reasoning fixes the count for cos⁡x=k\cos x=k and, after the RR-form, for asin⁡x+bcos⁡x=ca\sin x+b\cos x=c: that equation has no solution when ∣c∣>a2+b2|c|>\sqrt{a^2+b^2}, one (tangent) when ∣c∣=a2+b2|c|=\sqrt{a^2+b^2}, and two per period when ∣c∣<a2+b2|c|<\sqrt{a^2+b^2}.

Example
Worked example: parameter condition

For which values of kk does 3 sin⁡x+cos⁡x=k\sqrt3\,\sin x + \cos x = k have exactly one solution on each period? Rewrite in RR-form: R=(3)2+12=2R=\sqrt{(\sqrt3)^2+1^2}=2, so the equation is 2sin⁡(x+φ)=k2\sin(x+\varphi)=k, i.e. sin⁡(x+φ)=k2\sin(x+\varphi)=\tfrac{k}{2}. A basic sine equation has exactly one solution per period only at the tangent values k2=±1\tfrac{k}{2}=\pm1, giving

k=2  or  k=−2.\boxed{k = 2 \ \text{ or } \ k = -2.}

For ∣k∣<2|k|<2 there are two solutions per period; for ∣k∣>2|k|>2 there are none.

Concept
Systems of trig equations

To solve two trig equations simultaneously, find each variable's candidate set, then keep only the pairs that satisfy both. A frequent case is a system that hides an identity — for instance, given sin⁡x+sin⁡y\sin x + \sin y and cos⁡x+cos⁡y\cos x + \cos y, apply sum-to-product to both and divide to eliminate a variable. Always finish by checking candidates in both original equations, since combining equations (adding, dividing, squaring) can introduce extraneous pairs.

Example
Worked example: a small system

Solve the system on [0,2π)[0,2\pi):

sin⁡x+sin⁡y=1,x+y=π2.\sin x + \sin y = 1, \qquad x + y = \frac{\pi}{2}.

From the second equation y=π2−xy=\tfrac{\pi}{2}-x, so sin⁡y=sin⁡ ⁣(π2−x)=cos⁡x\sin y = \sin\!\big(\tfrac{\pi}{2}-x\big)=\cos x. Substitute:

sin⁡x+cos⁡x=1.\begin{aligned} \sin x + \cos x &= 1. \end{aligned}

Use the RR-form with a=b=1a=b=1: R=2R=\sqrt2, φ=π4\varphi=\tfrac{\pi}{4}, giving 2 sin⁡ ⁣(x+π4)=1\sqrt2\,\sin\!\big(x+\tfrac{\pi}{4}\big)=1, so sin⁡ ⁣(x+π4)=12\sin\!\big(x+\tfrac{\pi}{4}\big)=\tfrac{1}{\sqrt2}. Then x+π4=π4x+\tfrac{\pi}{4}=\tfrac{\pi}{4} or 3π4\tfrac{3\pi}{4}, i.e. x=0x=0 or x=π2x=\tfrac{\pi}{2}. The matching y=π2−xy=\tfrac{\pi}{2}-x gives

Solutions: (x,y)=(0, π2)  and  (π2, 0).\textbf{Solutions: } (x,y) = \Big(0,\ \tfrac{\pi}{2}\Big) \ \text{ and } \ \Big(\tfrac{\pi}{2},\ 0\Big).

Both check in the original system.

Tip

Convert before you count. Advanced equations rarely give up their solutions in raw form. First reshape — sum-to-product, double-angle, or the RR-form — until you reach sin⁡(angle)=c\sin(\text{angle})=c or a factored product. Only then apply the interval and period rules, and always re-check candidates from squaring, dividing, or combining equations against the original.

Try it
Quick check

How many solutions does cos⁡4x=−0.3\cos 4x=-0.3 have on [0,2π)[0,2\pi)? Answer without solving.

Show the answer

Answer: 88. As xx runs over [0,2π)[0,2\pi), 4x4x runs over [0,8π)[0,8\pi): four full turns. Since −1<−0.3<1-1<-0.3<1, each turn contains two angles with cosine −0.3-0.3, giving 4⋅2=84\cdot 2=8 solutions.

Why It Works

A trig equation has infinitely many solutions because the functions are periodic, and the structure of the solution set reflects the unit circle: a given value of sine occurs at two points on the circle (mirror images across the yy-axis), so sin⁡x=c\sin x = c has two solutions per period. The general solution x=θ+2πnx = \theta + 2\pi n or x=π−θ+2πnx = \pi - \theta + 2\pi n lists both families; for tangent only one family is needed, with period π\pi.

Tip
Reading a proof

Do not read a proof the way you read a rule. Cover the last line, ask what the previous line forces, and only then uncover it. If a step feels like magic, that is the step to work by hand with a concrete number.

Formulas, Proofs & Tips

Tip
Solving trigonometric equations
sin⁡θ=k⇒θ=θ0+2πn or π−θ0+2πn\sin\theta=k \Rightarrow \theta=\theta_0+2\pi n \ \text{or}\ \pi-\theta_0+2\pi n

What it means. Because trig functions repeat, every equation has infinitely many solutions.

Example. sin⁡θ=12\sin\theta=\tfrac12 on [0,2π)[0,2\pi) gives θ=π6\theta=\tfrac\pi6 and 5π6\tfrac{5\pi}6.

Why it works. On the unit circle a given sine value is the same height at two points per revolution — one at θ0\theta_0 and its mirror at π−θ0\pi-\theta_0 — and the whole picture repeats every 2π2\pi.

Tip. Find every solution in one period first, then add the multiples of the period. Check the interval the question actually asks for.