√2 cosx + cos2x = sin2x

Đáp án: $S=\{\ \dfrac{\pi}{4}+k\dfrac{2\pi}{3};\dfrac{3\pi}{4}+k2\pi |k\in Z \}$

Giải thích các bước giải:

 $\sqrt{2}cosx+cos2x=sin2x$

$sin2x-cos2x=\sqrt{2}cosx$

Chia cả 2 vế cho $\sqrt{1+1}=\sqrt{2}$ ta được  :

$\dfrac{1}{\sqrt{2}}sin2x-\dfrac{1}{\sqrt{2}}cos2x=cosx$

$sin(2x-\dfrac{\pi}{4})=sin(\dfrac{\pi}{2}-x)$

\(\left[ \begin{array}{l}2x-\dfrac{\pi}{4}=\dfrac{\pi}{2}-x+k2\pi\\2x-\dfrac{\pi}{4}=\dfrac{\pi}{2}+x+k2\pi\end{array} \right.\) 

\(\left[ \begin{array}{l}x=\dfrac{\pi}{4}+k\dfrac{2\pi}{3}\\x=\dfrac{3\pi}{4}+k2\pi\end{array} \right.,(k\in Z)\)

Vậy $S=\{\ \dfrac{\pi}{4}+k\dfrac{2\pi}{3};\dfrac{3\pi}{4}+k2\pi |k\in Z \}$