Question

Find all solutions of

Answer 1

Answer:
`x=n\pi +(-1)^n(\pi)/(6)`,
`x=n\pi +(-1)^n\left((-\pi)/(6)\right)`

Explanation:

Given

`\sin x-√(1-3\sin^2x)=0`

Take the square root term to the right-hand side

`\Rightarrow \sin x=√(1-3\sin^2x)\\\text{Squaring both sides}\\\Rightarrow \sin^2x=1-3\sin^2x\\\Rightarrow 4\sin^2x=1\\\\\Rightarrow \sin^2x=(1)/(4)\\\\\Rightarrow \sin x=\pm(1)/(2)`

for

`\sin x=(1)/(2)`

`x=n\pi +(-1)^n(\pi)/(6)`

for

`\sin x=(-1)/(2)`

`x=n\pi +(-1)^n\left((-\pi)/(6)\right)`