How to solve $\sin^2(x)+\sin2x+2\cos^2(x)=0$ How do you solve $\sin^2(x)+\sin2x+2\cos^2(x)=0$?
I have been able to rewrite it as $(\sin(x)+\cos(x))^2+\cos^2(x)=0$. Not obviously useful, I think
 A: Rushabh's answer is perfect, though I would do it slightly differently. Note that $\sin 2x = 2\sin x\cos x$, so your equation can be rewritten as
$$ \sin^2 x + 2\sin x\cos x + 2\cos^2 x = 0.$$
If you treat $\sin x$ and $\cos x$ as independent variables $u$ and $v$, this becomes the quadratic equation
$$ u^2 + 2uv + 2v^2 = 0, $$
where it becomes clear there are no solutions, as the discriminant is $\triangle = 2^2-4\times 2 = -4<0$.
A: From $$(\sin x + \cos x )^2 + \cos ^2x =0$$ we get  $$(\sin x + \cos x )^2=0$$ and $$\cos ^2x =0$$
The first identity implies $$\tan x =-1$$ and the second identity implies $$\cos x=0$$ which makes $\tan x = \pm \infty $.
Thus there is no solution which satisfies both identities, that is your equation has no solution.
A: $\sin^2(x) + \cos^2(x) = 1$, so we have:
$$\sin 2x + \cos^2 x + 1 = 0$$
Now, $\cos^2 x ≥ 0$, and $\sin 2x ≥ -1$, so the only solutions are when $\cos^2 x = 0$ and $\sin 2x = -1$.
When $\cos^2 x = 0$, $x = -\frac{\pi}{2} + 2\pi n$ or $x = \frac{\pi}{2} + 2\pi n$. However, $\sin \left( 2(-\frac{\pi}{2}) \right) = 0$ and $\sin \left( 2(\frac{\pi}{2}) \right) = 0$, so there are no solutions.
A: Hint:
For $a\sin^2x+b\cos^2x+c\sin2x=0,$
divide both sides by $\cos^2x$ as $\cos x\ne0$(why?)
to find $$at^2+2ct+b=0$$ where $t=\tan x$
For real solution, the discriminant must be $\ge0$
A: What you've done here is actually perfect! We know that the only way for this expression to be $0$ is if
$$\sin(x)+\cos(x)=0$$ and $$\cos^2(x)=0\to\cos(x)=0$$
We know both can't be possible because that would imply $\sin(x)=\cos(x)=0$, which is false. So the solution of given equation does not exist.
