giải phương trình
Câu 12,
$(1+tan^{2}x)(cosx+2)-sin^2x=cos^2x$
Câu 13,
$4cos^5x.sinx-4sin^5x.cosx=sin^24x$
Câu 14,
$4cos^3x+3\sqrt{2}sin2x=8cosx$
giải phương trình
Câu 12,
$(1+tan^{2}x)(cosx+2)-sin^2x=cos^2x$
Câu 13,
$4cos^5x.sinx-4sin^5x.cosx=sin^24x$
Câu 14,
$4cos^3x+3\sqrt{2}sin2x=8cosx$
Giải thích các bước giải:
$\begin{array}{l}
12)DK:\cos x \ne 0\\
\left( {1 + {{\tan }^2}x} \right)\left( {\cos x + 2} \right) – {\sin ^2}x = {\cos ^2}x\\
\Leftrightarrow \dfrac{1}{{{{\cos }^2}x}}\left( {\cos x + 2} \right) = 1\\
\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\left( c \right)\\
\cos x = 2\left( l \right)
\end{array} \right.\\
\Leftrightarrow x = \pi + k2\pi \left( {k \in Z} \right)\\
13)4{\cos ^5}x.\sin x – 4{\sin ^5}x.\cos x = {\sin ^2}4x\\
\Leftrightarrow 4\sin x\cos x\left( {{{\cos }^4}x – {{\sin }^4}x} \right) = {\sin ^2}4x\\
\Leftrightarrow 2\sin 2x\left( {{{\cos }^2}x – {{\sin }^2}x} \right) = {\sin ^2}4x\\
\Leftrightarrow 2\sin 2x\cos 2x = {\sin ^2}4x\\
\Leftrightarrow \sin 4x = {\sin ^2}4x\\
\Leftrightarrow \sin 4x\left( {\sin 4x – 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin 4x = 0\\
\sin 4x = 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = k\dfrac{\pi }{4}\\
x = \dfrac{\pi }{8} + k\dfrac{\pi }{2}
\end{array} \right.\left( {k \in Z} \right)\\
14)4{\cos ^3}x + 3\sqrt 2 \sin 2x = 8\cos x\\
\Leftrightarrow \left[ \begin{array}{l}
\cos x = 0\\
4{\cos ^2}x + 6\sqrt 2 \sin x = 8
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\cos x = 0\\
2{\cos ^2}x + 3\sqrt 2 \sin x = 4
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\cos x = 0\\
2\left( {1 – {{\sin }^2}x} \right) + 3\sqrt 2 \sin x = 4
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\cos x = 0\\
2{\sin ^2}x – 3\sqrt 2 \sin x + 2 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\cos x = 0\\
\left( {\sin x – \sqrt 2 } \right)\left( {\sin x – \dfrac{{\sqrt 2 }}{2}} \right) = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\cos x = 0\\
\sin x = \sqrt 2 \left( l \right)\\
\sin x = \dfrac{{\sqrt 2 }}{2}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\cos x = 0\\
\sin x = \dfrac{{\sqrt 2 }}{2}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k\pi \\
x = \dfrac{\pi }{4} + k2\pi \\
x = \dfrac{{3\pi }}{4} + k2\pi
\end{array} \right.\left( {k \in Z} \right)
\end{array}$