Đáp án:
\(\left[\begin{array}{l}x = \dfrac{\pi}{4} - \arccos\dfrac{4}{\sqrt{17}} + k\pi\\x = \dfrac{\pi}{4}+ k\pi\end{array}\right.\quad (k\in\Bbb Z)\)
Giải thích các bước giải:
\(\begin{array}{l}
\quad 2\cos2x + 4(\sin x + \cos x)^2 = 7 + 2\cos^2x\\
\Leftrightarrow 2\cos2x + 4(1 + \sin2x) = 7 + (1 + \cos2x)\\
\Leftrightarrow 4\sin2x + \cos2x = 4\\
\Leftrightarrow \dfrac{4}{\sqrt{17}}\sin2x + \dfrac{1}{\sqrt{17}}\cos2x = \dfrac{4}{\sqrt{17}}\\
\text{Đặt}\ \begin{cases}\cos\alpha = \dfrac{4}{\sqrt{17}}\\\sin\alpha = \dfrac{1}{\sqrt{17}}\end{cases}\Rightarrow \alpha = \arccos\dfrac{4}{\sqrt{17}}\\
\text{Phương trình trở thành:}\\
\quad \sin2x.\cos\alpha + \cos2x.\sin\alpha = \cos\alpha\\
\Leftrightarrow \sin(2x + \alpha) = \sin\left(\dfrac{\pi}{2} -\alpha\right)\\
\Leftrightarrow \left[\begin{array}{l}2x + \alpha =\dfrac{\pi}{2}- \alpha + k2\pi\\2x + \alpha = \dfrac{\pi}{2} + \alpha + k2\pi\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{4} - \alpha + k\pi\\x = \dfrac{\pi}{4}+ k\pi\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{4} - \arccos\dfrac{4}{\sqrt{17}} + k\pi\\x = \dfrac{\pi}{4}+ k\pi\end{array}\right.\quad (k\in\Bbb Z)\\
\text{Vậy phương trình có các họ nghiệm là}\ x = \dfrac{\pi}{4} - \arccos\dfrac{4}{\sqrt{17}} + k\pi\ \text{và}\ x = x = \dfrac{\pi}{4}+ k\pi\ \text{với}\ k\in\Bbb Z
\end{array}\)
