Đáp án:

 b) $x = \dfrac{\pi }{4} + k\dfrac{\pi }{2};x = \dfrac{1}{2}\arcsin \left( {\dfrac{3}{4}} \right) + k\pi ;x = \dfrac{\pi }{2} - \dfrac{1}{2}\arcsin \left( {\dfrac{3}{4}} \right) + k\pi \left( {k \in Z} \right)$

c) ${x = \dfrac{\pi }{2} + k\pi ;x = \dfrac{\pi }{4} + k\dfrac{\pi }{2};x = \dfrac{\pi }{6} + k2\pi ;x = \dfrac{{ - \pi }}{6} + k2\pi ;x = \dfrac{{5\pi }}{6} + k2\pi ;x = \dfrac{{ - 5\pi }}{6} + k2\pi \left( {k \in Z} \right)}$

Giải thích các bước giải:

$\begin{array}{l}
b)\sin 2x + \sin 6x = 3{\cos ^2}2x\\
 \Leftrightarrow \sin 2x + 3\sin 2x - 4{\sin ^3}2x - 3{\cos ^2}2x = 0\\
 \Leftrightarrow 4\sin 2x\left( {1 - {{\sin }^2}2x} \right) - 3{\cos ^2}2x = 0\\
 \Leftrightarrow 4\sin 2x{\cos ^2}2x - 3{\cos ^2}2x = 0\\
 \Leftrightarrow {\cos ^2}2x\left( {4\sin 2x - 3} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\cos 2x = 0\\
\sin 2x = \dfrac{3}{4}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
2x = \dfrac{\pi }{2} + k\pi \\
2x = \arcsin \left( {\dfrac{3}{4}} \right) + k2\pi \\
2x = \pi  - \arcsin \left( {\dfrac{3}{4}} \right) + k2\pi 
\end{array} \right.\left( {k \in Z} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{4} + k\dfrac{\pi }{2}\\
x = \dfrac{1}{2}\arcsin \left( {\dfrac{3}{4}} \right) + k\pi \\
x = \dfrac{\pi }{2} - \dfrac{1}{2}\arcsin \left( {\dfrac{3}{4}} \right) + k\pi 
\end{array} \right.\left( {k \in Z} \right)
\end{array}$

Vậy phương trình có các họ nghiệm là:

$x = \dfrac{\pi }{4} + k\dfrac{\pi }{2};x = \dfrac{1}{2}\arcsin \left( {\dfrac{3}{4}} \right) + k\pi ;x = \dfrac{\pi }{2} - \dfrac{1}{2}\arcsin \left( {\dfrac{3}{4}} \right) + k\pi \left( {k \in Z} \right)$

$\begin{array}{l}
c){\sin ^2}x = {\cos ^2}2x + {\cos ^2}3x\\
 \Leftrightarrow {\left( {2{{\cos }^2}x - 1} \right)^2} + {\left( {4{{\cos }^3}x - 3\cos x} \right)^2} - {\sin ^2}x = 0\\
 \Leftrightarrow {\left( {2{{\cos }^2}x - 1} \right)^2} + {\left( {4{{\cos }^3}x - 3\cos x} \right)^2} + {\cos ^2}x - 1 = 0\\
 \Leftrightarrow 16{\cos ^6}x - 20{\cos ^4}x + 6{\cos ^2}x = 0\\
 \Leftrightarrow 2{\cos ^2}x\left( {8{{\cos }^4}x - 10{{\cos }^2}x + 3} \right) = 0\\
 \Leftrightarrow 2{\cos ^2}x\left( {2{{\cos }^2}x - 1} \right)\left( {4{{\cos }^2} - 3} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\cos x = 0\\
\cos x = \dfrac{1}{{\sqrt 2 }}\\
\cos x = \dfrac{{ - 1}}{{\sqrt 2 }}\\
\cos x = \dfrac{{\sqrt 3 }}{2}\\
\cos x = \dfrac{{ - \sqrt 3 }}{2}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k\pi \\
x = \dfrac{\pi }{4} + k\dfrac{\pi }{2}\\
x = \dfrac{\pi }{6} + k2\pi \\
x = \dfrac{{ - \pi }}{6} + k2\pi \\
x = \dfrac{{5\pi }}{6} + k2\pi \\
x = \dfrac{{ - 5\pi }}{6} + k2\pi 
\end{array} \right.\left( {k \in Z} \right)
\end{array}$

Vậy các họ nghiệm của phương trình là: ${x = \dfrac{\pi }{2} + k\pi ;x = \dfrac{\pi }{4} + k\dfrac{\pi }{2};x = \dfrac{\pi }{6} + k2\pi ;x = \dfrac{{ - \pi }}{6} + k2\pi ;x = \dfrac{{5\pi }}{6} + k2\pi ;x = \dfrac{{ - 5\pi }}{6} + k2\pi \left( {k \in Z} \right)}$