√3(sin2x+cos7x)=sin7x-cos2x

Đáp án:

$\left\{\begin{array}{l}x=\dfrac{\pi}{10}+k\dfrac{2\pi}{5} \\x=\dfrac{7\pi}{54}+k\dfrac{2\pi}{9} \end{array} \right .(k\in\mathbb Z)$

Lời giải:

$\sqrt3(\sin2x+\cos7x)=\sin7x-\cos2x$

$\Leftrightarrow \sqrt3\sin2x+\cos2x=\sin7x-\sqrt3\cos7x$

$\Leftrightarrow \dfrac{\sqrt3}{2}\sin 2x+\dfrac{1}{2}\cos 2x=\dfrac{1}{2}\sin 7x-\dfrac{\sqrt3}{2}\cos 7x$

$\Leftrightarrow \sin(2x+\dfrac{\pi}{6})=\sin(7x-\dfrac{\pi}{3})$

$\Leftrightarrow \left[ \begin{array}{l} 2x+\dfrac{\pi}{6}=7x-\dfrac{\pi}{3}+k2\pi \\2x+\dfrac{\pi}{6}=\pi-7x+\dfrac{\pi}{3} +k2\pi\end{array} \right .$

$\Leftrightarrow \left[\begin{array}{l} x=\dfrac{\pi}{10}+k\dfrac{2\pi}{5} \\x=\dfrac{7\pi}{54}+k\dfrac{2\pi}{9} \end{array} \right .(k\in\mathbb Z)$

Vậy phương trình có nghiệm

$\left\{\begin{array}{l}x=\dfrac{\pi}{10}+k\dfrac{2\pi}{5} \\x=\dfrac{7\pi}{54}+k\dfrac{2\pi}{9} \end{array} \right .(k\in\mathbb Z)$

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