Đáp án:
Giải thích các bước giải:
$\text{ ta có}$
$sinx.cosx=$$\frac{1}{2}$ $sin2x$
giải
$2sin2xcos2x$$+\sqrt[]{3}$ $cos4x=$$-\sqrt[]{2}$
⇔$\frac{1}{2}$ $sin4x+$$\sqrt[]{3}$$cos4x=$ $\frac{\sqrt2}{2}$
⇔$sin4x+$$\sqrt[]{3}$ $cos4x=$ $\sqrt[]{2}$
⇔$2sin$$(4x+\frac{\pi}{3})=$ $\sqrt[]{2}$
⇔$sin(4x+\frac{\pi}{3})=$ $\frac{\sqrt2}{2}$
⇔ $sin(4x+\frac{\pi}{3})=$$sin\frac{\pi}{4}$
⇔\(\left[ \begin{array}{l}4x+\frac\pi{3}=\frac{\pi}{4}+k2\pi\\4x+\frac\pi{3}=\frac{3\pi}{4}+k2\pi\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=\frac{-\pi}{48}+k2\pi\\x=\frac{5\pi}{48}+k2\pi\end{array} \right.\)
