Tổng các nghiệm trên [- -$\pi$ ;$\pi$ ] của phương trình sin2x=cosx là?

`~rai~`

$sin2x=cosx$

$\Leftrightarrow sin2x-cosx=0$

$\Leftrightarrow 2sinxcosx-cosx=0$

$\Leftrightarrow cosx(sin2x-1)=0$

$\Leftrightarrow$\(\left[ \begin{array}{l}cosx=0\\2sinx-1=0\end{array} \right.\) 

$\Leftrightarrow$\(\left[ \begin{array}{l}x=\dfrac{\pi}{2}+k\pi\\sinx=\dfrac{1}{2}\end{array} \right.\) 

$\Leftrightarrow$\(\left[ \begin{array}{l}x_1=\dfrac{\pi}{2}+k\pi\\x_2=\dfrac{\pi}{6}+k2\pi(k\in\mathbb{Z})\\x_3=\dfrac{5\pi}{6}+k2\pi\end{array} \right.\) 

$+)TH1:-\pi \le x_1 \le \pi$

$\Leftrightarrow -\pi\le\dfrac{\pi}{2}+k\pi\le\pi$

$\Leftrightarrow-\dfrac{3\pi}{2}\le k\pi\le\dfrac{\pi}{2}$

$\Leftrightarrow-\dfrac{3}{2}\le k\le\dfrac{1}{2}$

$\text{Do k}\in\mathbb{Z}\Rightarrow k=-1;k=0$

$\Leftrightarrow x=-\dfrac{\pi}{2};x=\dfrac{\pi}{2}.$

$+)TH2:-\pi\le x_2\le\pi$

$\Leftrightarrow -\pi\le\dfrac{\pi}{6}+k2\pi\le\pi$

$\Leftrightarrow -\dfrac{7\pi}{6}\le k2\pi\le\dfrac{5\pi}{6}$

$\Leftrightarrow-\dfrac{7\pi}{12}\le k\le\dfrac{5\pi}{12}$

$\Leftrightarrow -\dfrac{7}{12}\le k\le\dfrac{5}{12}$

$\text{Do k}\in\mathbb{Z}\Rightarrow k=0$

$\Leftrightarrow x=\dfrac{\pi}{6}.$

$+)TH3:-\pi\le x_3\le\pi$

$\Leftrightarrow -\pi\le\dfrac{5\pi}{6}+k2\pi\le\pi$

$\Leftrightarrow -\dfrac{11\pi}{6}\le k2\pi\le\dfrac{\pi}{6}$

$\Leftrightarrow -\dfrac{11\pi}{12}\le k\pi\le\dfrac{\pi}{12}$

$\Leftrightarrow -\dfrac{11}{12}\le k\le\dfrac{1}{12}$

$\text{Do k}\in\mathbb{Z}\Rightarrow k=0$

$\Leftrightarrow x=\dfrac{5\pi}{6}.$

$\text{Tổng các nghiệm của phương trình là:}\dfrac{\pi}{2}-\dfrac{\pi}{2}+\dfrac{\pi}{6}+\dfrac{5\pi}{6}=\pi.$