Question

6 sin ^2x - cos x - 4 = 0

Answer 1

$6 sin ^2x - cos x - 4 = 0;, 6 -6 cos ^2x - cos x - 4 = 0;,$
$6 cos ^2x + cos x - 2 = 0;, cos x=t;, 6t^2+t-2=0;,D=49=7^2;$
$t_1= frac{-1+7}{12}= frac{1}{2};, t_2= frac{-1-7}{12}= -frac{2}{3};$
$, x_1=pm frac{ pi}{3}+2pi k,,kinmathbb{Z};, x_2=pm arccos( -frac{2}{3})+2pi n,,ninmathbb{Z}$