# Is $\sin^2\theta + \cos\theta = 2$ solvable without a mess?

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?

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@DavidMitra, Sorry, I meant $\sin^2\theta + \cos\theta = 2$. I'm not used to the math notation. – mowwwalker Apr 8 '12 at 2:16
The maximum of $\sin^2\theta+\cos\theta$ is $\frac54$ when $\theta$ is real, so something is definitely fishy... – Rahul Apr 8 '12 at 2:21
@RahulNarain, Thank you. I'm not sure if this was an intentional trick question or not on my teacher's part. – mowwwalker Apr 8 '12 at 2:24
Though the problem is already solved, let $\cos \theta =x$ and solve $x^2-x+1=0$. It has no solutions. – Pedro Tamaroff Apr 8 '12 at 2:30

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.

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Thank you. I'm not sure if this was an intentional trick question or not on my teacher's part. – mowwwalker Apr 8 '12 at 2:24
The equation has no real solutions. It has four complex solutions. $\approx\pm 1.196\pm 0.831\imath$ – Mark Adler Apr 8 '12 at 5:44
It can't have just four complex solutions. If $\theta$ is a solution, then so is $\theta+2\pi$. – Boris Bukh Apr 8 '12 at 9:00
Indeed, it is those solutions plus $2k\pi$, for any integer $k$. – Mark Adler Apr 8 '12 at 13:25
@MarkAdler: Yes, indeed. I was working under the assumption OP was looking for real solutions. I have edited to reflect this. – Dejan Govc Apr 8 '12 at 17:06

Almost there. Solve the quadratic equation and get $\cos\theta={1\pm \imath\sqrt{3}\over 2}$. Take $\pm\cos^{-1}({1\pm \imath\sqrt{3}\over 2})$ to get the answer. (It's $\pm$ since both $\cos$ and $\sin^2$ are even functions.) You can look up how to do complex $\cos^{-1}$ and $\log$ (you'll see why you need $\log$).

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+1. If we're going to catalog complex solutions, there are infinitely many. These are just the ones with real part between $-\pi$ and $\pi$. – alex.jordan Apr 8 '12 at 7:45
These are all of those. – Mark Adler Apr 8 '12 at 13:21
Are we miscommunicating? I mean that the full solution set is $\left\{ 2k\pi\pm\arccos \left(\frac{1\pm i\sqrt{3}}{2}\right)\mid k\in\mathbb{Z} \right\}$, where $\arccos$ is some branch of an inverse function for $\cos$. Er, maybe you are saying that too. – alex.jordan Apr 8 '12 at 19:29
Since you wrote "arccos", perhaps you are thinking of arccos as the library function in just about every computer language that returns a principal value, as opposed to an equation with multiple solutions. $\cos^{-1}(x)$ has multiple solutions, and isn't even a function unless you pick a branch. – Mark Adler Apr 8 '12 at 21:48
I disagree with your vocabulary in that last comment in three places, but it I'm sure we agree in concept. – alex.jordan Apr 9 '12 at 0:59

we have $|\sin[x]|\le 1$, thus $|\sin[x]|^{2}\le 1$. Your equation would imply both $\sin[x]$ and $\cos[x]$ has absolute value 1, which does not hold since then $\sin[2x]=2\sin[x]\cos[x]=\pm2$. Maybe you copied the wrong formula, etc.

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Maybe you copied the wrong formula, etc. Or maybe it was intended as an easy question... – GEdgar Apr 8 '12 at 17:09