Use the quadratic formula to solve trigonometric trinomial Solve $$-\sin^2\theta  + 2\cos\theta +\cos^2\theta = 0$$
Using the quadratic formula.
This is what you should get $$\theta = \cos^{-1}\biggl(\frac{-1+\sqrt{3}}{2}\biggr)$$
How do you set this up and solve?

i would write
$$-(1-\cos(^2(x))+2\cos(x)+\cos^2(x)=0$$
this gives $$2\cos^2(x)+2\cos(x)-1=0$$
$$\cos(x)^2+\cos(x)-\frac{1}{2}=0$$
Setting $t=\cos(x)$ we get
$$t_{1,2}=\frac{-1}{2}\pm\sqrt{\frac{1}{4}+\frac{1}{2}}$$
