Question

a. pi/3 b. pi/2c. 2pi/3d. 3pi/2 These are 4 options but there can be more than 2 or 3 correct answers.Find the solution of each equation the interval

Answer 1

$\sin x\cdot\cos x-\frac{\sqrt[]{3}}{2}\cos x=0$$\cos x(\sin x-\frac{\sqrt[]{3}}{2})=0$$\begin{gathered} \cos x=0 \\ x=\frac{\pi}{2},\frac{3\pi}{2} \end{gathered}$$\begin{gathered} \sin x-\frac{\sqrt[]{3}}{2}=0 \\ \sin x=\frac{\sqrt[]{3}}{2} \\ x=\frac{\pi}{3} \end{gathered}$

so the answer is pi/2, 3pi/2, and pi/3