Đáp án:
$ x=\dfrac{\alpha+k\pi}{2},cos\alpha=\dfrac{2}{\sqrt{5}}, sin\alpha =\dfrac{1}{\sqrt{5}}$
Giải thích các bước giải:
$2sin2x+2sin^2x-1=0$
$\rightarrow 2sin2x-(1-2sin^2x)=0$
$\rightarrow 2sin2x-cos2x=0$
Đặt $cos\alpha=\dfrac{2}{\sqrt{5}}, sin\alpha =\dfrac{1}{\sqrt{5}}$
$\rightarrow sin 2x.\dfrac{2}{\sqrt{5}}-cos2x.\dfrac{1}{\sqrt{5}}=0$
$\rightarrow sin2x.cos\alpha-cos2x.sin\alpha=0$
$\rightarrow sin(2x-\alpha)=0$
$\rightarrow 2x-\alpha=k\pi$
$\rightarrow x=\dfrac{\alpha+k\pi}{2}$
