Câu hỏi của Anh Trâm

Question

giải pt $cos\left(2x+\frac{2\pi}{3}\right)+3cos\left(x+\frac{\pi}{3}\right)+2=0$

Nguyễn Việt Lâm · Accepted Answer

$\Leftrightarrow2cos^2\left(x+\frac{\pi}{3}\right)-1+3cos\left(x+\frac{\pi}{3}\right)+2=0$

$\Leftrightarrow2cos^2\left(x+\frac{\pi}{3}\right)+3cos\left(x+\frac{\pi}{3}\right)+1=0$

$\Leftrightarrow\left[{}\begin{matrix}cos\left(x+\frac{\pi}{3}\right)=-1\cos\left(x+\frac{\pi}{3}\right)=-\frac{1}{2}\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{3}=\pi+k2\pi\x+\frac{\pi}{3}=\frac{2\pi}{3}+k2\pi\x+\frac{\pi}{3}=-\frac{2\pi}{3}+k2\pi\end{matrix}\right.$

$\Leftrightarrow...$Câu trả lời: $\Leftrightarrow2cos^2\left(x+\frac{\pi}{3}\right)-1+3cos\left(x+\frac{\pi}{3}\right)+2=0$

$\Leftrightarrow2cos^2\left(x+\frac{\pi}{3}\right)+3cos\left(x+\frac{\pi}{3}\right)+1=0$

$\Leftrightarrow\left[{}\begin{matrix}cos\left(x+\frac{\pi}{3}\right)=-1\cos\left(x+\frac{\pi}{3}\right)=-\frac{1}{2}\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{3}=\pi+k2\pi\x+\frac{\pi}{3}=\frac{2\pi}{3}+k2\pi\x+\frac{\pi}{3}=-\frac{2\pi}{3}+k2\pi\end{matrix}\right.$

$\Leftrightarrow...$

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