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

Step-by-step explanation:

Given

[tex]\sin x-\sqrt{1-3\sin^2x}=0[/tex]

Take the square root term to the right-hand side

[tex]\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}[/tex]

for

[tex]\sin x=\dfrac{1}{2}[/tex]

[tex]x=n\pi +(-1)^n\dfrac{\pi}{6}[/tex]

for

[tex]\sin x=\dfrac{-1}{2}[/tex]

[tex]x=n\pi +(-1)^n\left(\dfrac{-\pi}{6}\right)[/tex]

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