Đáp án:
Vậy phương trình có các họ nghiệm là:
a) ${x = k2\pi \left( {k \in Z} \right);x = \dfrac{{2\pi }}{3} + k2\pi \left( {k \in Z} \right);x = - \dfrac{{2\pi }}{3} + k2\pi \left( {k \in Z} \right)}$
b) $x = \dfrac{\pi }{2} + k2\pi \left( {k \in Z} \right)$
c) $x = - \dfrac{\pi }{{24}} + k\dfrac{\pi }{2}\left( {k \in Z} \right)$
Giải thích các bước giải:
$\begin{array}{l}
a)2{\cos ^2}\dfrac{x}{2} + \cos \dfrac{x}{2} - 1 = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\cos x = 1\\
\cos x = \dfrac{{ - 1}}{2}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = k2\pi \\
x = \dfrac{{2\pi }}{3} + k2\pi \\
x = - \dfrac{{2\pi }}{3} + k2\pi
\end{array} \right.\\
2)\cos 2x - \sin x + 2 = 0\\
\Leftrightarrow 1 - 2{\sin ^2}x - \sin x + 2 = 0\\
\Leftrightarrow 2{\sin ^2}x + \sin x - 3 = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x = 1\\
\sin x = \dfrac{{ - 3}}{2}\left( l \right)
\end{array} \right.\\
\Leftrightarrow \sin x = 1\\
\Leftrightarrow x = \dfrac{\pi }{2} + k2\pi \\
c)\sqrt 3 \cos 4x - \sin 4x = 2\\
\Leftrightarrow \dfrac{{\sqrt 3 }}{2}\cos 4x - \dfrac{1}{2}\sin 4x = 1\\
\Leftrightarrow \cos \left( {4x + \dfrac{\pi }{6}} \right) = 1\\
\Leftrightarrow 4x + \dfrac{\pi }{6} = k2\pi \\
\Leftrightarrow x = - \dfrac{\pi }{{24}} + k\dfrac{\pi }{2}
\end{array}$
