Đáp án:

\(\begin{array}{l}
a)\,\,Pt\,\,\,vo\,\,\,nghiem.\\
b)\,\,\left[ \begin{array}{l}
x = \frac{\pi }{{16}} + \frac{{k\pi }}{2}\\
x = \frac{\pi }{8} + \frac{{k\pi }}{3}
\end{array} \right.\,\,\,\left( {k \in Z} \right).\\
c)\,\,\,Pt\,\,\,vo\,\,\,nghiem.\\
d)\,\,\left[ \begin{array}{l}
x = \frac{{k\pi }}{3}\\
x = \frac{{k\pi }}{7}
\end{array} \right.\,\,\,\,\left( {k \in Z} \right).
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\,\,2{\sin ^2}x + \cos x - 3 = 0\\
 \Leftrightarrow 2\left( {1 - {{\cos }^2}x} \right) + \cos x - 3 = 0\\
 \Leftrightarrow 2{\cos ^2}x - \cos x + 1 = 0\\
 \Rightarrow Pt\,\,\,vo\,\,\,nghiem.\\
b)\,\,\,\sin x + \cos x = \sqrt 2 \sin 5x\\
 \Leftrightarrow \sqrt 2 \sin \left( {x + \frac{\pi }{4}} \right) = \sqrt 2 \sin 5x\\
 \Leftrightarrow \sin \left( {x + \frac{\pi }{4}} \right) = \sin 5x\\
 \Leftrightarrow \left[ \begin{array}{l}
x + \frac{\pi }{4} = 5x + k2\pi \\
x + \frac{\pi }{4} = \pi  - 5x + k2\pi 
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \frac{\pi }{{16}} + \frac{{k\pi }}{2}\\
x = \frac{\pi }{8} + \frac{{k\pi }}{3}
\end{array} \right.\,\,\,\left( {k \in Z} \right).\\
c)\,\,5{\sin ^2}x - 3\sin x.\cos x + {\cos ^2}x = 0\,\,\,\,\left( * \right)\\
Voi\,\,\,\cos x = 0 \Rightarrow \left( * \right) \Leftrightarrow 5{\sin ^2}x = 0\,\,\,\left( {vo\,\,\,ly} \right)\\
 \Rightarrow \cos x = 0\,\,\,\,khong\,\,la\,\,nghiem\,\,\,cua\,\,\left( * \right)\\
Chia\,\,\,ca\,\,\,2\,\,\,ve\,\,\,cua\,\,\,pt\,\,\,\left( * \right)\,\,\,cho\,\,\,{\cos ^2}\,x\,\,\,ta\,\,duoc:\\
\left( * \right) \Leftrightarrow 5{\tan ^2}x - 3\tan x + 1 = 0\\
 \Rightarrow pt\,\,vo\,\,\,nghiem.\\
d)\,\,{\cos ^2}2x - {\sin ^2}2x = \sin \left( {10,5\pi  - 10x} \right)\\
 \Leftrightarrow \cos 4x = \sin \left( {10\pi  + \frac{\pi }{2} - 10x} \right)\\
 \Leftrightarrow \cos 4x = \cos 10x\\
 \Leftrightarrow \left[ \begin{array}{l}
4x = 10x + k2\pi \\
4x =  - 10x + k2\pi 
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \frac{{k\pi }}{3}\\
x = \frac{{k\pi }}{7}
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)
\end{array}\)