Đáp án:
\(\begin{array}{l}
4,\\
x = \pi + k2\pi \,\,\,\,\left( {k \in Z} \right)\\
6,\\
x = \pi + k2\pi \,\,\,\,\left( {k \in Z} \right)
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
4,\\
2{\sin ^2}x + \cos x + {\cos ^2}x = 0\\
\Leftrightarrow 2.\left( {1 - {{\cos }^2}x} \right) + \cos x + {\cos ^2}x = 0\\
\Leftrightarrow 2 - 2{\cos ^2}x + \cos x + {\cos ^2}x = 0\\
\Leftrightarrow - {\cos ^2}x + \cos x + 2 = 0\\
\Leftrightarrow {\cos ^2}x - \cos x - 2 = 0\\
\Leftrightarrow \left( {\cos x + 1} \right)\left( {\cos x - 2} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\cos x + 1 = 0\\
\cos x - 2 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\cos x = - 1\\
\cos x = 2
\end{array} \right.\\
- 1 \le \cos x \le 1 \Rightarrow \cos x = - 1 \Leftrightarrow x = \pi + k2\pi \,\,\,\,\left( {k \in Z} \right)\\
6,\\
5{\sin ^2}x + 3\cos x + 3 = 0\\
\Leftrightarrow 5.\left( {1 - {{\cos }^2}x} \right) + 3\cos x + 3 = 0\\
\Leftrightarrow 5 - 5{\cos ^2}x + 3\cos x + 3 = 0\\
\Leftrightarrow - 5{\cos ^2}x + 3\cos x + 8 = 0\\
\Leftrightarrow 5{\cos ^2}x - 3\cos x - 8 = 0\\
\Leftrightarrow \left( {\cos x + 1} \right)\left( {5\cos x - 8} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\cos x + 1 = 0\\
5\cos x - 8 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\cos x = - 1\\
\cos x = \dfrac{8}{5}
\end{array} \right.\\
- 1 \le \cos x \le 1 \Rightarrow \cos x = - 1 \Leftrightarrow x = \pi + k2\pi \,\,\,\,\left( {k \in Z} \right)
\end{array}\)
