Đáp án`+`Giải thích các bước giải:

 `sin^2 (2x) + cos^2 x = 1`

`<=>1 - sin^2 (2x) - cos^2 x = 0`

`<=> cos^2 (2x) - cos^2 x= 0`

`<=> (cos (2x) - cosx)(cos (2x) + cosx) = 0`

`<=>` $\left[\begin{matrix} cosx = cos(2x)\\ cosx = -cos (2x)\end{matrix}\right.$ 

`<=>` $\left[\begin{matrix} cosx = cos (2x)\\cosx =cos(\pi -2x)\end{matrix}\right.$

`<=>` $\left[\begin{matrix} x= 2x + k2\pi\\x = -2x + k2\pi\\ x = \pi -2x + k2\pi\\x = 2x - \pi + k2\pi\end{matrix}\right.$ `(k \in ZZ)`

`<=>`$\left[\begin{matrix} x=-k2\pi\\ x= \dfrac{k2\pi}{3}\\x = \dfrac{\pi + k2\pi}{3}\\ x = \pi - k2\pi \end{matrix}\right.$ `(k \in ZZ)`

`<=>` $\left[\begin{matrix} x=-k2\pi\\ x= \dfrac{k2\pi}{3}\\x = \dfrac{\pi + k2\pi}{3}\\ x= \pi - k2\pi\end{matrix}\right.$ `(k \in ZZ)`

Vậy `S= {-k2\pi; (k2\pi)/3;(\pi + k2\pi)/3; \pi -k2\pi}   (k \in ZZ)`

`***`sharksosad