Đáp án:
Giải thích các bước giải:
$\sqrt[]{3}sinx+cosx=$ $\sqrt[]{2}$
⇔$\frac{\sqrt3}{2}sinx+$ $\frac{1}{2}cosx=$ $\sqrt[]{2}$
⇔$sinx^{}.cos$ $\frac{\pi}{6}+cosx.sin$ $\frac{\pi}{6}=sin$ $\frac{\pi}{4}$
⇔$sin(x^{}+\frac{\pi}{6})=sin\frac{\pi}{4}$
⇔\(\left[ \begin{array}{l}x+\frac\pi{6}=\frac\pi{4}+k2\pi\\x+\frac{\pi}{6}=\frac{3\pi}{4}+k2\pi\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=\frac{\pi}{12}+k2\pi\\x=\frac{7\pi}{12}+k2\pi\end{array} \right.\)
