Đáp án:
\(\begin{array}{l}
a,\\
\left[ \begin{array}{l}
x = \dfrac{{3\pi }}{4} + k2\pi \\
x = - \dfrac{{3\pi }}{4} + k2\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
b,\\
x = k\pi \,\,\,\,\left( {k \in Z} \right)\\
c,\\
\left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k2\pi \\
x = - \dfrac{\pi }{6} + k2\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\cos x = \cos y \Leftrightarrow \left[ \begin{array}{l}
x = y + k2\pi \\
x = - y + k2\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
\sin x = \sin y \Leftrightarrow \left[ \begin{array}{l}
x = y + k2\pi \\
x = \pi - y + k2\pi
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)\\
a,\\
\cos x = - \dfrac{{\sqrt 2 }}{2}\\
\Leftrightarrow \cos x = \cos \dfrac{{3\pi }}{4}\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{3\pi }}{4} + k2\pi \\
x = - \dfrac{{3\pi }}{4} + k2\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
b,\\
{\sin ^2}x - 3\sin x = 0\\
\Leftrightarrow \sin x.\left( {\sin x - 3} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x = 0\\
\sin x - 3 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\sin x = 0\\
\sin x = 3
\end{array} \right.\,\,\\
- 1 \le \sin x \le 1 \Rightarrow \sin x = 0 \Leftrightarrow x = k\pi \,\,\,\,\left( {k \in Z} \right)\\
c,\\
\sin x + \sqrt 3 \cos x = 1\\
\Leftrightarrow \dfrac{{\sqrt 3 }}{2}\cos x + \dfrac{1}{2}\sin x = \dfrac{1}{2}\\
\Leftrightarrow \cos x.\cos \dfrac{\pi }{6} + \sin x.\sin \dfrac{\pi }{6} = \dfrac{1}{2}\\
\Leftrightarrow \cos \left( {x - \dfrac{\pi }{6}} \right) = \cos \dfrac{\pi }{3}\\
\Leftrightarrow \left[ \begin{array}{l}
x - \dfrac{\pi }{6} = \dfrac{\pi }{3} + k2\pi \\
x - \dfrac{\pi }{6} = - \dfrac{\pi }{3} + k2\pi
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k2\pi \\
x = - \dfrac{\pi }{6} + k2\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)
\end{array}\)
