substitute:
cos(2x)=cos²(x)-sin²(x)
=2cos²(x)-1
to get
8cos²(x)+2cos(x)-1=0
let y=cos(x),
8y²+2y-1=0
(4y-1)(2y+1)=0
y=1/4, y=-1/2
cos(x)=1/4, cos(x)=-1/2
therefore
x=2kπ±acos(1/4), or
x=2kπ±acos(-1/2)
You may want to express the solutions in a more familiar form.
4cos(2x)+2cos(x)+3=0
1 answer