[Math] Is $\sin^2\theta + \cos\theta = 2$ solvable without a mess

algebra-precalculustrigonometry

I'm given the problem $\sin^2\theta + \cos\theta = 2$ and I'm told to use the pythagorean identity $\sin^2\theta + \cos^2\theta = 1$ to solve it.

I end up with $\cos^2\theta – \cos\theta + 1 = 0$, but I know that's not going to factor and solve very nicely.

Did I do something wrong, or is the answer going to end up being very ugly?

Best Answer

The equation has no real solutions.

For every $\theta\in\mathbb R$, we have $\sin^2\theta\in[0,1]$ and $\cos\theta\in[-1,1]$. This means that $\sin^2\theta+\cos\theta=2$ is only possible if $\sin^2\theta=1$ and $\cos\theta=1$. But if $\sin^2\theta=1$ we immediately have $\cos^2\theta=1-\sin^2\theta=0$, so $\cos\theta$ would have to be equal to $0$. This means the equation has no real solutions.