Question

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Answer 1

Answer: $x=n\pi +(-1)^n\dfrac{\pi}{6}$, $x=n\pi +(-1)^n\left(\dfrac{-\pi}{6}\right)$

Step-by-step explanation:

Given

$\sin x-\sqrt{1-3\sin^2x}=0$

Take the square root term to the right-hand side

$\Rightarrow \sin x=\sqrt{1-3\sin^2x}\\\text{Squaring both sides}\\\Rightarrow \sin^2x=1-3\sin^2x\\\Rightarrow 4\sin^2x=1\\\\\Rightarrow \sin^2x=\dfrac{1}{4}\\\\\Rightarrow \sin x=\pm\frac{1}{2}$

for

$\sin x=\dfrac{1}{2}$

$x=n\pi +(-1)^n\dfrac{\pi}{6}$

for

$\sin x=\dfrac{-1}{2}$

$x=n\pi +(-1)^n\left(\dfrac{-\pi}{6}\right)$