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?
 A: Following the comment on your question, we note that, since $\sin^2\theta+\cos^2\theta=1$, we can write $-\sin^2\theta=\cos^2\theta-1$. Thus, the equation becomes:
$$\cos^2\theta-1+2\cos\theta+\cos^2\theta=0$$
or:
$$2X^2+2X-1=0$$
for $X=\cos\theta$. Can you take it from there?
A: As @Mathmore said in the comments, substitute $\sin^2\theta=1-\cos^2\theta$ to obtain
$$-(1-\cos^2\theta)+2\cos \theta+\cos^2 \theta=0$$
$$2\cos^2\theta+2\cos \theta-1=0$$
Substitute $t=\cos \theta$. This is a quadratic equation 
$$2t^2+2t-1=0$$
with solutions
$$t=\frac{1}{2}(-1\pm \sqrt 3)$$
Note however that $t=\cos \theta \in[-1,1]$, that is $t\neq1/2(-1-\sqrt3)=-1.366\not\in[-1,1]$
So the solutions are
$$\cos \theta=\frac{1}{2}(\sqrt 3-1)$$
$$\theta=\pm\arccos(\frac{1}{2}(\sqrt 3-1))+2\pi n \quad n\in\mathbb{Z}$$
A: 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
$$t_{1,2}=\frac{-1}{2}\pm\sqrt{\frac{1}{4}+\frac{1}{2}}$$
