\(\text { Giải phương trình }: \frac{2\left(\cos ^{4} x-\sin ^{4} x\right)+1}{2 \cos \left(\frac{x}{2}-\frac{\pi}{3}\right)}=\sqrt{3} \cos x+\sin x\)
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Lời giải:
Báo sai\(\text { Đk: } x \neq \frac{5 \pi}{3}+k 2 \pi, k \in \mathbb{Z} \text { . Khi đó }\)
\(\begin{aligned} \operatorname{Pt}& \Leftrightarrow 2 \cos ^{2} x-2 \sin ^{2} x+1=2 \cos \left(\frac{x}{2}-\frac{\pi}{3}\right)(\sqrt{3} \cos x+\sin x) \\ & \Leftrightarrow 3 \cos ^{2} x-\sin ^{2} x=2 \cos \left(\frac{x}{2}-\frac{\pi}{3}\right)(\sqrt{3} \cos x+\sin x) \\ & \Leftrightarrow(\sqrt{3} \cos x+\sin x)(\sqrt{3} \cos x-\sin x)=2 \cos \left(\frac{x}{2}-\frac{\pi}{3}\right)(\sqrt{3} \cos x+\sin x) \\ & \Leftrightarrow\left[\begin{array} { l } { \sqrt { 3 } \operatorname { c o s } x + \operatorname { s i n } x = 0 } \\ { \sqrt { 3 } \operatorname { c o s } x - \operatorname { s i n } x = 2 \operatorname { c o s } ( \frac { x } { 2 } - \frac { \pi } { 3 } ) } \end{array} \Leftrightarrow \left[\begin{array}{l} \cos \left(x-\frac{\pi}{6}\right)=0 \\ \cos \left(x+\frac{\pi}{6}\right)=\cos \left(\frac{x}{2}-\frac{\pi}{3}\right) \end{array}\right.\right. \end{aligned}\)
\(\Leftrightarrow\left[\begin{array}{l} x=\frac{2 \pi}{3}+k \pi \\ x=-\pi+k 4 \pi \quad(k \in \mathbb{Z}) \\ x=\frac{\pi}{9}+k \frac{4 \pi}{3} \end{array}\right.\)
