sin^2 2x - 2cos^2 x + 3/4=0
6sin^2 3x + cos12x =14
1) [tex]pt\Leftrightarrow \frac{1-cos4x}{2}-(1+cos2x)+\frac{3}{4}=0[/tex]
Thay $cos4x=2cos^22x-1$ ra pt bậc 2
2) [tex]pt\Leftrightarrow 3(1-cos6x)+cos12x=14[/tex]
Thay $cos12x=2cos^26x-1$. Giống câu 1