Đáp án:
\[x = \pm 4\arccos \dfrac{{ - 1 + \sqrt {17} }}{4} + k8\pi \,\,\,\,\left( {k \in Z} \right)\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\cos \dfrac{x}{4} - 2{\sin ^2}\dfrac{x}{4} = 0\\
\Leftrightarrow \cos \dfrac{x}{4} - 2\left( {1 - {{\cos }^2}\dfrac{x}{4}} \right) = 0\\
\Leftrightarrow \cos \dfrac{x}{4} - 2 + 2{\cos ^2}\dfrac{x}{4} = 0\\
\Leftrightarrow 2{\cos ^2}\dfrac{x}{4} + \cos \dfrac{x}{4} - 2 = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\cos \dfrac{x}{4} = \dfrac{{ - 1 + \sqrt {17} }}{4}\\
\cos \dfrac{x}{4} = \dfrac{{ - 1 - \sqrt {17} }}{4}
\end{array} \right.\\
- 1 \le \cos \dfrac{x}{4} \le 1 \Rightarrow \cos \dfrac{x}{4} = \dfrac{{ - 1 + \sqrt {17} }}{4}\\
\cos \dfrac{x}{4} = \dfrac{{ - 1 + \sqrt {17} }}{4}\\
\Leftrightarrow \dfrac{x}{4} = \pm \arccos \dfrac{{ - 1 + \sqrt {17} }}{4} + k2\pi \\
\Leftrightarrow x = \pm 4\arccos \dfrac{{ - 1 + \sqrt {17} }}{4} + k8\pi \,\,\,\,\left( {k \in Z} \right)
\end{array}\)
