<s>[INST] How do I solve the trigonometric equation $1 - \sin^2x - \cos(2x) = \frac{1}{2}$? Solve for $x$ when $1-\sin^2x - \cos 2x = \dfrac{1}{2}$.
I can' t change it into a form I can work with. It is rather complicated.
 [/INST] As given $$1-\sin^2 x-\cos 2x=\frac{1}{2} \implies \cos^2x-\left(2\cos^2 x-1\right)=\frac{1}{2}$$ $$ \cos^2 x=\frac{1}{2}=\cos^2 \frac{\pi}{4} \implies x=n\pi\pm \frac{\pi}{4}$$ Hence, the general solution of the equation is $x=n\pi\pm \frac{\pi}{4} $ where, $n$ is an integer. 
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