Câu hỏi của Thảo Nguyễn Phương

Question

cos2x-√3 sin2x=sin3x+1

3sin2x+4cos2x+5cos2003x=0

√3sin(x-$\frac{\pi}{3}$)$+sin\left(x+\frac{\pi}{6}\right)-2sin1972x=0$

\(\sqrt{2}cos\left(\frac{x}{5}-\frac{\pi}{12}\right)-\sqrt{6}sin\left(\fr...

Nguyễn Việt Lâm · Accepted Answer

d/

$\Leftrightarrow\sqrt{2}\left(\frac{1}{2}cos\left(\frac{x}{5}-\frac{\pi}{12}\right)-\frac{\sqrt{3}}{2}sin\left(\frac{x}{5}-\frac{\pi}{12}\right)\right)=sin\left(\frac{x}{5}+\frac{2\pi}{3}\right)-sin\left(\frac{3x}{5}+\frac{\pi}{6}\right)$

$\Leftrightarrow\sqrt{2}cos\left(\frac{x}{5}-\frac{\pi}{12}+\frac{\pi}{3}\right)=2cos\left(\frac{2x}{5}+\frac{5\pi}{12}\right)sin\left(\frac{\pi}{4}-\frac{x}{5}\right)$

$\Leftrightarrow cos\left(\frac{x}{5}-\frac{\pi}{4}\right)=\sqrt{2}cos\left(\frac{2x}{5}+\frac{5\pi}{12}\right)cos\left(\frac{x}{5}-\frac{\pi}{4}\right)$

$\Leftrightarrow\left[{}\begin{matrix}cos\left(\frac{x}{5}-\frac{\pi}{4}\right)=0\cos\left(\frac{2x}{5}+\frac{5\pi}{12}\right)=\frac{\sqrt{2}}{2}\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}\frac{x}{5}-\frac{\pi}{4}=\frac{\pi}{2}+k\pi\\frac{2x}{5}+\frac{5\pi}{12}=\frac{\pi}{4}+k2\pi\\frac{2x}{5}+\frac{5\pi}{12}=-\frac{\pi}{4}+k2\pi\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x=\frac{15\pi}{4}+k5\pi\x=-\frac{5\pi}{12}+k5\pi\x=-\frac{5\pi}{3}+k5\pi\end{matrix}\right.$Câu trả lời: d/

$\Leftrightarrow\sqrt{2}\left(\frac{1}{2}cos\left(\frac{x}{5}-\frac{\pi}{12}\right)-\frac{\sqrt{3}}{2}sin\left(\frac{x}{5}-\frac{\pi}{12}\right)\right)=sin\left(\frac{x}{5}+\frac{2\pi}{3}\right)-sin\left(\frac{3x}{5}+\frac{\pi}{6}\right)$

$\Leftrightarrow\sqrt{2}cos\left(\frac{x}{5}-\frac{\pi}{12}+\frac{\pi}{3}\right)=2cos\left(\frac{2x}{5}+\frac{5\pi}{12}\right)sin\left(\frac{\pi}{4}-\frac{x}{5}\right)$

$\Leftrightarrow cos\left(\frac{x}{5}-\frac{\pi}{4}\right)=\sqrt{2}cos\left(\frac{2x}{5}+\frac{5\pi}{12}\right)cos\left(\frac{x}{5}-\frac{\pi}{4}\right)$

$\Leftrightarrow\left[{}\begin{matrix}cos\left(\frac{x}{5}-\frac{\pi}{4}\right)=0\cos\left(\frac{2x}{5}+\frac{5\pi}{12}\right)=\frac{\sqrt{2}}{2}\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}\frac{x}{5}-\frac{\pi}{4}=\frac{\pi}{2}+k\pi\\frac{2x}{5}+\frac{5\pi}{12}=\frac{\pi}{4}+k2\pi\\frac{2x}{5}+\frac{5\pi}{12}=-\frac{\pi}{4}+k2\pi\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x=\frac{15\pi}{4}+k5\pi\x=-\frac{5\pi}{12}+k5\pi\x=-\frac{5\pi}{3}+k5\pi\end{matrix}\right.$

Nguyễn Việt Lâm · Answer

c/

$\Leftrightarrow\sqrt{3}sin\left(x-\frac{\pi}{3}\right)+cos\left(\frac{\pi}{3}-x\right)=2sin1972x$

$\Leftrightarrow\frac{\sqrt{3}}{2}sin\left(x-\frac{\pi}{3}\right)+\frac{1}{2}cos\left(x-\frac{\pi}{3}\right)=sin1972x$

$\Leftrightarrow sin\left(x-\frac{\pi}{3}+\frac{\pi}{6}\right)=sin1972x$

$\Leftrightarrow sin\left(x-\frac{\pi}{6}\right)=sin1972x$

$\Leftrightarrow\left[{}\begin{matrix}1972x=x-\frac{\pi}{6}+k2\pi\1972x=\frac{7\pi}{6}-x+k2\pi\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{11826}+\frac{k2\pi}{1971}\x=\frac{7\pi}{11838}+\frac{k2\pi}{1973}\end{matrix}\right.$Câu trả lời: c/

$\Leftrightarrow\sqrt{3}sin\left(x-\frac{\pi}{3}\right)+cos\left(\frac{\pi}{3}-x\right)=2sin1972x$

$\Leftrightarrow\frac{\sqrt{3}}{2}sin\left(x-\frac{\pi}{3}\right)+\frac{1}{2}cos\left(x-\frac{\pi}{3}\right)=sin1972x$

$\Leftrightarrow sin\left(x-\frac{\pi}{3}+\frac{\pi}{6}\right)=sin1972x$

$\Leftrightarrow sin\left(x-\frac{\pi}{6}\right)=sin1972x$

$\Leftrightarrow\left[{}\begin{matrix}1972x=x-\frac{\pi}{6}+k2\pi\1972x=\frac{7\pi}{6}-x+k2\pi\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{11826}+\frac{k2\pi}{1971}\x=\frac{7\pi}{11838}+\frac{k2\pi}{1973}\end{matrix}\right.$

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