Question

8sin^2 x-14cos x +1=o
СРОЧНО НУЖНА ПОМОЩЬ

Answer 1

$8(1-cos^2x)-14cosx+1=0 \\cosx=y,\ y \in [-1;1] \\8-8y^2-14y+1=0 \\-8y^2-14y+9=0 \\8y^2+14y-9=0 \\D=196+288=484=22^2 \\y_1=\frac{-14+22}{16}=\frac{8}{16}=\frac{1}{2} \\y_2=\frac{-14-22}{16} \notin [-1;1] \\cosx=\frac{1}{2} \\x=\pm \frac{\pi}{3}+2\pi n,\ n \in Z$