# $\cos^2(x)+\cos(x)\sqrt{\cos^2(x)+3}+\sin(x)\cos(x)+\sin(x)\sqrt{\cos^2(x)+3}=\frac98$

So this equation: $$\cos^2(x)+\cos(x)\sqrt{\cos^2(x)+3}+\sin(x)\cos(x)+\sin(x)\sqrt{\cos^2(x)+3}=\frac98$$ is kind of like the final part of a very much bigger geometry problem, what I mean is that the equation was originally this big of mess

$$\frac{(4\cos(x)+\sqrt{16\cos^2(x)+48})^2-84+8\sin(x)(4\cos(x)+\sqrt{16\cos^2(x)+48})}{4}=0$$

so uh yeah I got there after simplifying. And I've also tried factoring but ended up with $$(\cos(x)+\sin(x))(\cos(x)+\sqrt{\cos^2(x)+3})-\frac98=0$$ and I'm kind of stuck, I was just wondering if it exists a way of doing this algebraically and not graphically. And yes, I know there will be a better method to the geometry problem and I know, but I'm interested in solving the equation, not about the geometry problem.

Hint:

As $$\sqrt{\cos^2x+3}-\cos x=\dfrac3{\sqrt{\cos^2x+3}+\cos x}$$

$$8(\cos x+\sin x)=3(\sqrt{\cos^2x+3}-\cos x)$$

$$\iff8\sin x+11\cos x=3\sqrt{\cos^2x+3}$$

As $$\cos x\ne0,$$ divide both sides by $$\cos x$$

$$8\tan x+11=3\sqrt{3+1+\tan^2x}$$

Now take square both sides to form a quadratic equation in $$\tan x$$ and check for When do we get extraneous roots?