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