sinx=2cosx^2

Question

solve for 0 degrees< X <  180 degrees

Answer 1

sin(x)=2cos²(x)=2(1-sin²(x))=2-2sin²(x),

2sin²(x)+sin(x)=2,

sin²(x)+½sin(x)=1,

sin²(x)+½sin(x)+1/16=1+1/16 (completing the square),

(sin(x)+¼)²=17/16,

sin(x)+¼=±√17/4, sin(x)=-¼+√17/4=0.7808 or -¼-√17/4=-1.2808 (approximately).

We can ignore the negative value because it's less than -1 and sine is always between -1 and +1.

So sin(x)=0.7808, x=51.33° or 180-51.33=128.67°.