Đáp án:
c) \(\left[ \begin{array}{l}
x = - \dfrac{\pi }{{12}} + k\pi \\
x = \dfrac{{7\pi }}{{12}} + k\pi
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a){\sin ^2}x + {\cos ^2}x = 1\left( {ld} \right)\forall x\\
b){\sin ^2}3x + {\sin ^2}x = 1\\
\to {\left( {3\sin x - 4{{\sin }^3}x} \right)^2} + {\sin ^2}x = 1\\
\to 9{\sin ^2}x - 24{\sin ^4}x + 16{\sin ^6}x + {\sin ^2}x = 1\\
\to 16{\sin ^6}x - 24{\sin ^4} + 10{\sin ^2}x = 1\\
Đặt:{\sin ^2}x = t\left( {t > 0} \right)\\
Pt \to 16{t^3} - 24{t^2} + 10t - 1 = 0\\
\to \left[ \begin{array}{l}
t = \dfrac{{2 - \sqrt 2 }}{4}\\
t = \dfrac{{2 + \sqrt 2 }}{4}\\
t = \dfrac{1}{2}
\end{array} \right.\\
\to \left[ \begin{array}{l}
{\sin ^2}x = \dfrac{{2 - \sqrt 2 }}{4}\\
{\sin ^2}x = \dfrac{{2 + \sqrt 2 }}{4}\\
{\sin ^2}x = \dfrac{1}{2}
\end{array} \right.\\
\to \left[ \begin{array}{l}
\sin x = \pm \sqrt {\dfrac{{2 - \sqrt 2 }}{4}} \\
\sin x = \pm \sqrt {\dfrac{{2 + \sqrt 2 }}{4}} \\
\sin x = \pm \sqrt {\dfrac{1}{2}}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \arcsin \left( { \pm \sqrt {\dfrac{{2 - \sqrt 2 }}{4}} } \right)\\
x = \arcsin \left( { \pm \sqrt {\dfrac{{2 + \sqrt 2 }}{4}} } \right)\\
x = \arcsin \left( { \pm \sqrt {\dfrac{1}{2}} } \right)
\end{array} \right.\\
c)4\sin x.\cos x + 1 = 0\\
\to 2\sin 2x + 1 = 0\\
\to \sin 2x = - \dfrac{1}{2}\\
\to \left[ \begin{array}{l}
2x = - \dfrac{\pi }{6} + k2\pi \\
2x = \dfrac{{7\pi }}{6} + k2\pi
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - \dfrac{\pi }{{12}} + k\pi \\
x = \dfrac{{7\pi }}{{12}} + k\pi
\end{array} \right.\left( {k \in Z} \right)
\end{array}\)
