Question

please help me solve this, I got: pi/2, 4pi/3, and 5pi/3. maybe i'm missing a solution or didn't eliminate one properly? because it was wrong. thank you for any help!

Answer 1

The expression is given as

$\sin 2x=\sqrt[]{3}\text{ cosx}$

Simplify the expression by using the identity

sin2x=2 sinx cosx.

Substitute the identity in the expression above

$2\sin x\text{ cosx=}\sqrt[]{3}\cos x$$2\sin x\cos x-\sqrt[]{3}\cos x=0$$\cos x(2\sin x-\sqrt[]{3})=0$$\cos x=0,\sin x=\frac{\sqrt[]{3}}{2}$

For cosx =0,

$x=\frac{\pi}{2},\frac{3\pi}{2}$

For

$\sin x=\frac{\sqrt[]{3}}{2}$$x=\frac{\pi}{3},\frac{2\pi}{3}$

Hence the solution of x for the given interval [0,2pi) is

$x=\frac{\pi}{2},\frac{3\pi}{2},\frac{\pi}{3},\frac{2\pi}{3}$