Ta có
$\dfrac{\sin x\cos x}{4\sin^{2} x+7\cos^{2} x+3\cos 2x}=\dfrac{2\sin x\cos x}{8\sin^{2} x+14\cos^{2} x+6\cos 2x}=\dfrac{\sin2 x}{4(1-\cos2x)+7(1+\cos2x)+6\cos 2x}$
$=\dfrac{\sin2 x}{9\cos 2x+11}=-\dfrac{1}{18}.\dfrac{(9\cos 2x+11)'}{9\cos 2x+11}$
Suy ra 
 $\int\limits_{}^{} \dfrac{\sin x\cos x}{4\sin^{2} x+7\cos^{2} x+3\cos 2x}dx=-\dfrac{1}{18}\ln|9\cos 2x+11|+C$

PT
$\Leftrightarrow (9\sin x-9) + (6\cos x - 3\sin 2x) + \cos 2x +1= 0$
$\Leftrightarrow 9(\sin x-1) + 6\cos x(1-\sin x) +2(1-\sin^2 x)= 0$
 $\Leftrightarrow (\sin x-1)(9 - 6\cos x -2-2\sin x)= 0$
 $\Leftrightarrow (\sin x-1)(7 - 6\cos x -2\sin x)= 0$
 Mặt khác ta thấy $6^2 +2^2 <7^2$ nên PT $7 = 6\cos x +2\sin x$ vô nghiệm.
Do vậy $\sin x= 1 \Leftrightarrow x = \dfrac{\pi}{2}+k2\pi, k \in \mathbb Z.$

b) Đặt $t=-x\Rightarrow dt=-dx$
$I= \int\limits_{-\pi/4}^{\pi/4} \frac{\sin^{6}x + \cos^{6}x}{6^{x}+1}dx= \int\limits_{\pi/4}^{-\pi/4} \frac{\sin^{6}(-t) + \cos^{6}(-t)}{6^{-t}+1}(-dt)= \int\limits_{-\pi/4}^{\pi/4} \frac{(\sin^{6}t + \cos^{6}t)6^t}{1+6^{t}}dt$
$\implies I= \int\limits_{-\pi/4}^{\pi/4} \frac{(\sin^{6}x + \cos^{6}x)6^x}{1+6^{x}}dx$
Suy ra
$2I=I+I = \int\limits_{-\pi/4}^{\pi/4} \frac{\sin^{6}x + \cos^{6}x}{6^{x}+1}dx+ \int\limits_{-\pi/4}^{\pi/4} \frac{(\sin^{6}x + \cos^{6}x)6^x}{1+6^{x}}dx= \int\limits_{-\pi/4}^{\pi/4}(\sin^{6}x + \cos^{6}x)dx$
$= \int\limits_{-\pi/4}^{\pi/4}(\dfrac{5}{8}+\dfrac{3}{8}\cos 4x)dx=\left[ {\dfrac{5x}{8}+\dfrac{3}{32}\sin 4x} \right]_{-\pi/4}^{\pi/4}=\dfrac{5\pi}{16}$

a)
$\cos^{2}3x.\cos^{2}6x=\dfrac{1+\cos 6x}{2}.\dfrac{1+\cos 12x}{2}=\dfrac{1}{4}+\dfrac{1}{4}\cos 6x + \dfrac{1}{4}\cos 12x+\dfrac{1}{4}\cos6x\cos12x$
$=\dfrac{1}{4}+\dfrac{1}{4}\cos 6x + \dfrac{1}{4}\cos 12x+\dfrac{1}{8}\cos 18x +\dfrac{1}{8}\cos6x$
$=\dfrac{1}{4}+\dfrac{3}{8}\cos 6x + \dfrac{1}{4}\cos 12x+\dfrac{1}{8}\cos 18x $
 Suy ra
$ \int\limits_{0}^{\pi/2} (\cos^{2}3x.\cos^{2}6x)dx=\left[ {\dfrac{x}{4}+\dfrac{1}{16}\sin 6x + \dfrac{1}{48}\sin 12x+\dfrac{1}{144}\sin 18x} \right]_{0}^{\pi/2}= \dfrac{\pi}{8}$