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=(\pi)/(2),(3\pi)/(2)`

For

`\sin x=\frac{\sqrt[]{3}}{2}`
`x=(\pi)/(3),(2\pi)/(3)`

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

`x=(\pi)/(2),(3\pi)/(2),(\pi)/(3),(2\pi)/(3)`