Question 112354
*[Tex \LARGE 3\sin^2(x)=\cos^2(x)] Start with the given equation



*[Tex \LARGE 3\sin^2(x)-\cos^2(x)=0] Subtract *[Tex \Large \cos^2(x)] from both sides



*[Tex \LARGE 3\sin^2(x)-(1-\sin^2(x))=0] Replace *[Tex \Large \cos^2(x)] with *[Tex \Large 1-\sin^2(x)]


*[Tex \LARGE 3\sin^2(x)-1+\sin^2(x)=0] Distribute the negative



*[Tex \LARGE 4\sin^2(x)-1=0] Combine like terms


*[Tex \LARGE 4\sin^2(x)=1] Add 1 to both sides



*[Tex \LARGE \sin^2(x)=\frac{1}{4}] Divide both sides by 4



*[Tex \LARGE \sin(x)=sqrt{\frac{1}{4}}] Take the square root of both sides



*[Tex \LARGE \sin(x)=\frac{1}{2}] Simplify



*[Tex \LARGE arcsin(\sin(x))=arcsin(\frac{1}{2})] Take the arcsine of both sides



*[Tex \LARGE x=\frac{\pi}{6}] Take the arcsine of both sides



Since *[Tex \LARGE 0\le x \le \pi], the only solution is 



*[Tex \LARGE x=\frac{\pi}{6}] or x=30 degrees