I have such an exercise:
$$\color{teal}{{|x|\over{x}}\sin^2x-\cos|x|\cos x=1} $$ What I did is so:
If $x\ge 0$ then we have:
$$\sin^2x-\cos^2x=1$$
$$\sin^2x=1$$
So: $$\sin x=1$$ or $$\sin x=-1$$
This means that we have two sets of solutions: $\color{blue}{x={\pi\over2}+2\pi k}$ and $\color{blue}{x={3\pi \over 2}+2\pi k}$
The problem is I do not know what solutions to choose from the set I presented above, knowing that $x\ge0$.
Now, it's your turn if you please. What values should be choosen? My opinion is that $k\ge 0$ Thank you very much!:)
Hints:
$$x\ge 0\implies \sin^2x-\cos^2x=-\cos2x=1\iff \cos 2x=-\frac12\iff\ldots$$
$$x<0\implies -\sin^2x-\cos^2x=-(\sin^2x+\cos^2x)=-1\neq1\;\ldots$$