Câu hỏi của Trần Khánh Huyền

Question

$\frac{sin\left(2x+\frac{3\pi}{4}\right)}{c\text{os}\left(x+\frac{\pi}{4}\right)}+1=0$

Nguyễn Việt Lâm · Accepted Answer

ĐKXĐ: ...

$\Leftrightarrow sin\left(2x+\frac{3\pi}{4}\right)+cos\left(x+\frac{\pi}{4}\right)=0$

$\Leftrightarrow cos\left(x+\frac{\pi}{4}\right)=-sin\left(2x+\frac{3\pi}{4}\right)$

$\Leftrightarrow cos\left(x+\frac{\pi}{4}\right)=cos\left(2x+\frac{5\pi}{4}\right)$

$\Rightarrow\left[{}\begin{matrix}2x+\frac{5\pi}{4}=x+\frac{\pi}{4}+k2\pi\2x+\frac{5\pi}{4}=-x-\frac{\pi}{4}+k2\pi\end{matrix}\right.$

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

$\Leftrightarrow sin\left(2x+\frac{3\pi}{4}\right)+cos\left(x+\frac{\pi}{4}\right)=0$

$\Leftrightarrow cos\left(x+\frac{\pi}{4}\right)=-sin\left(2x+\frac{3\pi}{4}\right)$

$\Leftrightarrow cos\left(x+\frac{\pi}{4}\right)=cos\left(2x+\frac{5\pi}{4}\right)$

$\Rightarrow\left[{}\begin{matrix}2x+\frac{5\pi}{4}=x+\frac{\pi}{4}+k2\pi\2x+\frac{5\pi}{4}=-x-\frac{\pi}{4}+k2\pi\end{matrix}\right.$

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

Chương 1: HÀM SỐ LƯỢNG GIÁC. PHƯƠNG TRÌNH LƯỢNG GIÁC