TrigonometryHardTrigonometric Equations ShareTrigonometric Equations · Hard problemSolve sinx+cosx=1\sin x+\cos x=1sinx+cosx=1 on [0,2π)[0,2\pi)[0,2π). (Square, then check for extraneous solutions.)Ax=0, π2x=0,\ \dfrac{\pi}{2}x=0, 2πBx=0, π3x=0,\ \dfrac{\pi}{3}x=0, 3πCx=0, π1x=0,\ \dfrac{\pi}{1}x=0, 1πDx=0, 2πx=0,\ \dfrac{2}{\pi}x=0, π2Check Show answer Walk through it step by step Practice more like this Review this concept