TrigonometryHardTrigonometric Equations ShareTrigonometric Equations · Hard problemSolve 2cos2x−5cosx+2=02\cos^2 x-5\cos x+2=02cos2x−5cosx+2=0 on [0,2π)[0,2\pi)[0,2π).Ax=3π, 5π3x=\dfrac{3}{\pi},\ \dfrac{5\pi}{3}x=π3, 35πBx=π6, 5π3x=\dfrac{\pi}{6},\ \dfrac{5\pi}{3}x=6π, 35πCx=π3, 5π4x=\dfrac{\pi}{3},\ \dfrac{5\pi}{4}x=3π, 45πDx=π3, 5π3x=\dfrac{\pi}{3},\ \dfrac{5\pi}{3}x=3π, 35πCheck Show answer Walk through it step by step Practice more like this Review this concept