Let's first of all convert everything to sinxsinx.
Consider the pythagorean identity sin^2x + cos^2x = 1sin2x+cos2x=1. Rearranging:
cos^2x = 1 - sin^2xcos2x=1−sin2x
Now substituting into our equation:
2(1 - sin^2x) - sinx - 1 = 02(1−sin2x)−sinx−1=0
2 - 2sin^2x - sinx - 1 =0 2−2sin2x−sinx−1=0
-2sin^2x - sinx + 1 = 0−2sin2x−sinx+1=0
-2sin^2x - 2sinx + sinx + 1 = 0−2sin2x−2sinx+sinx+1=0
-2sinx(sinx + 1) + 1(sinx + 1) = 0−2sinx(sinx+1)+1(sinx+1)=0
(-2sinx + 1)(sinx + 1) = 0(−2sinx+1)(sinx+1)=0
sinx = 1/2 and sinx = -1sinx=12andsinx=−1
x = pi/6, (5pi)/6, (3pi)/2x=π6,5π6,3π2
Note that these solutions are only in the interval 0 < x < 2pi0<x<2π.
Hopefully this helps!