Đáp án:
b. \(\left[ \begin{array}{l}
x = - \dfrac{\pi }{6} + k\pi \\
x = \dfrac{\pi }{2} + k\pi
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a.2{\cos ^2}x + 5\sin x - 4 = 0\\
\to 2\left( {1 - {{\sin }^2}x} \right) + 5\sin x - 4 = 0\\
\to - 2{\sin ^2}x + 5\sin x - 2 = 0\\
\to \left( {2 - \sin x} \right)\left( {2\sin x - 1} \right) = 0\\
\to \left[ \begin{array}{l}
\sin x = 2\left( l \right)\\
\sin x = \dfrac{1}{2}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{\pi }{6} + k2\pi \\
x = \dfrac{{5\pi }}{6} + k2\pi
\end{array} \right.\left( {k \in Z} \right)\\
b.{\sin ^4}x - {\cos ^4}x = 2\sqrt 3 \sin x.\cos x + 1\\
\to \left( {{{\sin }^2}x - {{\cos }^2}x} \right)\left( {{{\sin }^2}x + {{\cos }^2}x} \right) = \sqrt 3 \sin 2x + 1\\
\to - \left( {{{\cos }^2}x - {{\sin }^2}x} \right) = \sqrt 3 \sin 2x + 1\\
\to - \cos 2x = \sqrt 3 \sin 2x + 1\\
\to \sqrt 3 \sin 2x + \cos 2x = - 1\\
\to \dfrac{{\sqrt 3 }}{2}\sin 2x + \dfrac{1}{2}\cos 2x = - \dfrac{1}{2}\\
\to \sin 2x.\cos \dfrac{\pi }{6} + \sin \dfrac{\pi }{6}.\cos 2x = - \dfrac{1}{2}\\
\to \sin \left( {2x + \dfrac{\pi }{6}} \right) = - \dfrac{1}{2}\\
\to \left[ \begin{array}{l}
2x + \dfrac{\pi }{6} = - \dfrac{\pi }{6} + k2\pi \\
2x + \dfrac{\pi }{6} = \dfrac{{7\pi }}{6} + k2\pi
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - \dfrac{\pi }{6} + k\pi \\
x = \dfrac{\pi }{2} + k\pi
\end{array} \right.\left( {k \in Z} \right)
\end{array}\)
