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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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  • $\begingroup$ @DavidMitra, Sorry, I meant $\sin^2\theta + \cos\theta = 2$. I'm not used to the math notation. $\endgroup$
    – mowwwalker
    Apr 8, 2012 at 2:16
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    $\begingroup$ The maximum of $\sin^2\theta+\cos\theta$ is $\frac54$ when $\theta$ is real, so something is definitely fishy... $\endgroup$
    – user856
    Apr 8, 2012 at 2:21
  • $\begingroup$ @RahulNarain, Thank you. I'm not sure if this was an intentional trick question or not on my teacher's part. $\endgroup$
    – mowwwalker
    Apr 8, 2012 at 2:24
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    $\begingroup$ Though the problem is already solved, let $\cos \theta =x$ and solve $x^2-x+1=0$. It has no solutions. $\endgroup$
    – Pedro
    Apr 8, 2012 at 2:30

5 Answers 5

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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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    $\begingroup$ Thank you. I'm not sure if this was an intentional trick question or not on my teacher's part. $\endgroup$
    – mowwwalker
    Apr 8, 2012 at 2:24
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    $\begingroup$ The equation has no real solutions. It has four complex solutions. $\approx\pm 1.196\pm 0.831\imath$ $\endgroup$
    – Mark Adler
    Apr 8, 2012 at 5:44
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    $\begingroup$ It can't have just four complex solutions. If $\theta$ is a solution, then so is $\theta+2\pi$. $\endgroup$
    – Boris Bukh
    Apr 8, 2012 at 9:00
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    $\begingroup$ Indeed, it is those solutions plus $2k\pi$, for any integer $k$. $\endgroup$
    – Mark Adler
    Apr 8, 2012 at 13:25
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    $\begingroup$ @MarkAdler: Yes, indeed. I was working under the assumption OP was looking for real solutions. I have edited to reflect this. $\endgroup$
    – Dejan Govc
    Apr 8, 2012 at 17:06
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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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  • $\begingroup$ +1. If we're going to catalog complex solutions, there are infinitely many. These are just the ones with real part between $-\pi$ and $\pi$. $\endgroup$
    – 2'5 9'2
    Apr 8, 2012 at 7:45
  • $\begingroup$ These are all of those. $\endgroup$
    – Mark Adler
    Apr 8, 2012 at 13:21
  • $\begingroup$ 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. $\endgroup$
    – 2'5 9'2
    Apr 8, 2012 at 19:29
  • $\begingroup$ 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. $\endgroup$
    – Mark Adler
    Apr 8, 2012 at 21:48
  • $\begingroup$ I disagree with your vocabulary in that last comment in three places, but it I'm sure we agree in concept. $\endgroup$
    – 2'5 9'2
    Apr 9, 2012 at 0:59
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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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  • $\begingroup$ Maybe you copied the wrong formula, etc. Or maybe it was intended as an easy question... $\endgroup$
    – GEdgar
    Apr 8, 2012 at 17:09
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Because cosine is a wave that has a domain of all real numbers, you can plug in multiple different x's for the same y's. When you're solving this, remember that cosθ can give you different values.

Start by using the Pythagorean identity (which you did) and get cos^2θ−cosθ+1=2 (which is slightly off on the post). Move the 1 to the left, then factor out cosine: -cosθ(cosθ-1)=1

Now, set both parts to equal 1: -cosθ=1 cosθ-1=1

you get that cosθ = -1 and cosθ = 0

Now, use the inverse to find what θ is!

cos^-1(0) = θ and cos^-1(-1) = θ

I got pi and pi/2, hopefully that's right!

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    $\begingroup$ Welcome to MathSE. Your answer is incorrect (your factorization is wrong and you cannot use the zero product property unless one side is equal to zero). Please read the previous answers. For future reference, this tutorial explains how to typeset mathematics on this site. $\endgroup$ Jul 29, 2022 at 1:31
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Another way, by tangent half-angle formula

$$\sin^2\theta + \cos\theta = 2 \iff \frac{4t^2}{(1+t^2)^2}+\frac{1-t^2}{1+t^2}=2$$

$$\iff 4t^2+1-t^4=2+4t^2+2t^4 \iff 3t^4=-1$$

which is impossible.

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