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what is the answer of

2sin^2(x)-cos(2x)=0 ?

Verified Answer :

LammettHash

\(2\sin^2x-\cos2x=2\sin^2x-(2\cos^2x-1)=2(\sin^2-\cos^2x)+1=0\)
\(-2\cos2x=-1\)
\(\cos2x=\dfrac12\)

You have \(\cos x=\dfrac12\) for

\(x=\dfrac\pi3+2n\pi\)
\(x=-\dfrac\pi3+2n\pi\)

which means \(\cos 2x=\dfrac12\) for

\(2x=\dfrac\pi3+2n\pi\implies x=\dfrac\pi6+n\pi\)
\(2x=-\dfrac\pi3+2n\pi\implies x=-\dfrac\pi6+n\pi\)

where \(n\) is any integer.

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