1)
$6\sin^23x+\cos(2.6x)-14=0$
$\to 6.\dfrac{1-\cos6x}{2}+2\cos^26x-1-14=0$
$\to -3\cos6x+3+2\cos^26x-15=0$
$\to 2\cos^26x-3\cos6x-12=0$
$\to \cos6x=\dfrac{3\pm\sqrt{105}}{4}\notin [-1;1]$ (Loại)
Vậy PT vô nghiệm
2)
$4(\sin^2x)^2+12(1-\sin^2x)-7=0$
$\to 4(\sin^2x)^2-12\sin^2x+5=0$
$\to \left[ \begin{array}{l}\sin^2x=\dfrac{5}{2}>1(L) \\ \sin^2x=\dfrac{1}{2} \end{array} \right.$
$\to \dfrac{1-\cos2x}{2}=\dfrac{1}{2}$
$\to \cos2x=0$
$\to x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}$
3)
$5\sin3x+\cos(2.3x)+2=0$
$\to 5\sin3x+1-2\sin^23x+2=0$
$\to -2\sin^23x+5\sin3x+3=0$
$\to \left[ \begin{array}{l}\sin3x= 3>1(L) \\ \sin3x=\dfrac{-1}{2} \end{array} \right.$
$\to \left[ \begin{array}{l}3x=\dfrac{-\pi}{6}+k2\pi \\ 3x=\dfrac{7\pi}{6}+k2\pi \end{array} \right.$
$\to \left[ \begin{array}{l}x=\dfrac{-\pi}{18}+\dfrac{k2\pi}{3} \\x=\dfrac{7\pi}{18}+\dfrac{k2\pi}{3} \end{array} \right.$
4)
$4.(\sin^23x)^2+12(1-\sin^23x)-7=0$
$\to 4(\sin^23x)^2-12\sin^23x+5=0$
$\to \left[ \begin{array}{l}\sin^23x=\dfrac{5}{2}>1(L) \\\sin^23x=\dfrac{1}{2}\end{array} \right.$
$\to \dfrac{1-\cos6x}{2}=\dfrac{1}{2}$
$\to \cos6x=0$
$\to 6x=\dfrac{\pi}{2}+k\pi$
$\to x=\dfrac{\pi}{12}+\dfrac{k\pi}{6}$
