\[\begin{array}{l}
a)4{\sin ^2}2x + 8{\cos ^2}2x - 8 = 0 \\
\Leftrightarrow 4\left( {1 - {{\cos }^2}2x} \right) + 8{\cos ^2}2x - 8 = 0 \\
\Leftrightarrow 4{\cos ^2}2x = 4 \Leftrightarrow {\cos ^2}2x = 1 \\
\Leftrightarrow \left[ \begin{array}{l}
\cos 2x = 1 \\
\cos 2x = - 1 \\
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
2x = k2\Pi \\
2x = \Pi + k2\Pi \\
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = k\Pi \\
x = \dfrac{\Pi }{2} + k\Pi \\
\end{array} \right.;k \in Z \\
\end{array}\]
\[b)\left( {1 + {{\tan }^2}x} \right)\left( {\cos x + 2} \right) - {\sin ^2}x = {\cos ^2}x\]
Điều kiện
\[{\cos ^2}x \ne 0 \Leftrightarrow x \ne \dfrac{\Pi }{2} + k\Pi ,k \in Z\]
Phương trình \[\begin{array}{l}
\Leftrightarrow \dfrac{1}{{{{\cos }^2}x}}\left( {\cos x + 2} \right) - \left( {1 - {{\cos }^2}x} \right) = {\cos ^2}x \\
\Leftrightarrow - {\cos ^2}x + \cos x + 2 = 0 \\
\end{array}\]
Đặt \[t = \cos x\] với \[t \in \left[ { - 1,1} \right]\]
\[\begin{array}{l}
\Leftrightarrow - {t^2} + t + 2 = 0 \Leftrightarrow \left[ \begin{array}{l}
t = - 1 \to nhan \\
t = 2 \to loai \\
\end{array} \right. \\
\Rightarrow \cos x = - 1 \Leftrightarrow x = \Pi + k2\Pi ,k \in Z \to thoamandieukien \\
\end{array}\]
\[c)\dfrac{3}{{{{\cos }^2}x}} - 4\tan x - 2 = 0\]
Điều kiện
\[\cos x \ne 0 \Leftrightarrow x \ne \dfrac{\Pi }{2} + k\Pi ,k \in Z\]
\[\begin{array}{l}
\Leftrightarrow 3\left( {1 + {{\tan }^2}x} \right) - 4\tan x - 2 = 0 \\
\Leftrightarrow 3{\tan ^2}x - 4\tan x + 1 = 0 \\
\end{array}\]
Đặt \[t = \tan x\] với \[t \in R\]
Phương trình \[\begin{array}{l}
\Leftrightarrow 3{t^2} - 4t + 1 = 0 \Leftrightarrow \left[ \begin{array}{l}
t = 1 \\
t = \dfrac{1}{3} \\
\end{array} \right. \\
t = 1 \Rightarrow \tan x = 1 \Leftrightarrow x = \dfrac{\Pi }{4} + k\Pi ,k \in Z \to thoamandieukien \\
t = \dfrac{1}{3} \Rightarrow \tan x = \dfrac{1}{3} \Leftrightarrow x = \arctan \left( {\dfrac{1}{3}} \right) + k\Pi ,k \in Z \to thoamandieukien \\
\left[ \begin{array}{l}
x = \dfrac{\Pi }{4} + k\Pi \\
x = \arctan \left( {\dfrac{1}{3}} \right) + k\Pi \\
\end{array} \right.,k \in Z \\
\end{array}\]
\[d)5\cos 2x - 12\sin 2x = 13\]
Chia 2 vế cho
\[\begin{array}{l}
\sqrt {{5^2} + {{12}^2}} = 13 \\
\dfrac{5}{{13}}\cos 2x - \dfrac{{12}}{{13}}\sin 2x = 1 \\
\sin \alpha = \dfrac{5}{{13}},\cos \alpha = \dfrac{{12}}{{13}} \\
\Leftrightarrow \sin \alpha \cos 2x - \cos \alpha \sin 2x = 1 \\
\Leftrightarrow \sin \left( {\alpha - 2x} \right) = 1 \Leftrightarrow \alpha - 2x = \dfrac{\Pi }{2} + k2\Pi \\
\Leftrightarrow x = \dfrac{\alpha }{2} - \dfrac{\Pi }{4} + k\Pi ,k \in Z \\
\end{array}\]
\[e)\sin 7x + \sqrt 3 \cos 7x = \sqrt 2 \]
Chia 2 vế cho
\[\begin{array}{l}
\sqrt {{1^2} + {{\left( {\sqrt 3 } \right)}^2}} = 2 \\
\Leftrightarrow \dfrac{1}{2}\sin 7x + \dfrac{{\sqrt 3 }}{2}\cos 7x = \dfrac{{\sqrt 2 }}{2} \\
\Leftrightarrow \cos \dfrac{\Pi }{3}\sin 7x + \sin \dfrac{\Pi }{3}\cos 7x = \dfrac{{\sqrt 2 }}{2} \\
\Leftrightarrow \sin \left( {7x + \dfrac{\Pi }{3}} \right) = \sin \left( {\dfrac{\Pi }{4}} \right) \\
\Leftrightarrow \left[ \begin{array}{l}
7x + \dfrac{\Pi }{3} = \dfrac{\Pi }{4} + k2\Pi \\
7x + \dfrac{\Pi }{3} = \Pi - \dfrac{\Pi }{4} + k2\Pi \\
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = - \dfrac{\Pi }{{84}} + k\dfrac{{2\Pi }}{7} \\
x = \dfrac{5}{{84}} + k\dfrac{{2\Pi }}{7} \\
\end{array} \right.;k \in Z \\
\end{array}\]
\[f)\sin 4x + 4\cos 4x = \dfrac{{\sqrt {51} }}{2}\]
Chia 2 vế cho
\[\begin{array}{l}
\sqrt {{1^2} + {4^2}} = \sqrt {17} \\
\Leftrightarrow \dfrac{1}{{\sqrt {17} }}\sin 4x + \dfrac{4}{{\sqrt {17} }}c{\rm{os}}4x = \dfrac{{\sqrt 3 }}{2} \\
\end{array}\]
Đặt
\[\begin{array}{l}
c{\rm{os}}\alpha = \dfrac{1}{{\sqrt {17} }},\sin \alpha = \dfrac{4}{{\sqrt {17} }} \\
\Leftrightarrow c{\rm{os}}\alpha \sin 4x + \sin \alpha c{\rm{os}}4x = \dfrac{{\sqrt 3 }}{2} \\
\Leftrightarrow \sin \left( {4x + \alpha } \right) = \dfrac{{\sqrt 3 }}{2} \\
\Leftrightarrow \sin \left( {4x + \alpha } \right) = \sin \dfrac{\Pi }{3} \\
\Leftrightarrow \left[ \begin{array}{l}
4x + \alpha = \dfrac{\Pi }{3} + k2\Pi \\
4x + \alpha = \Pi - \dfrac{\Pi }{3} + k2\Pi \\
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = - \dfrac{\alpha }{4} + \dfrac{\Pi }{{12}} + k\dfrac{\Pi }{2} \\
x = - \dfrac{\alpha }{4} + \dfrac{\Pi }{6} + k\dfrac{\Pi }{2} \\
\end{array} \right.;k \in Z \\
\end{array}\]
Đơn giản thế là xong
