Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Solve the following for $\theta$:

$\cos^2 \theta + \cos \theta = 2$ [Hint: There is only one solution.]

I started this out by changing $\cos^2\theta$ to $\dfrac{1+\cos(2\theta)}{2}+\cos\theta=2$
$1+\cos(2\theta)$ turns into $1+\cos^2\theta-\sin^2\theta$ which all becomes; $\dfrac{1+\cos^2\theta-\sin^2\theta}{2}+\cos\theta=2$
Not to sure what to do after this. I was going to try a power reducing rule for $\sin^2\theta$ but that would make $\dfrac{1+\cos^2\theta- \left(\frac{1-\cos(2\theta)}2 \right)}2+\cos\theta=2$. Please do help.

share|cite|improve this question
Hint: $\cos x$ is bounded in absolute value by 1; so the equality holds only when $\cos x=1$. – David Mitra Jul 14 '12 at 23:50
This equation looks like a quadratic polynomial, with $cos\theta$ as the variable... – Francis Adams Jul 14 '12 at 23:52
up vote 4 down vote accepted

Replacing $\cos^2\theta$ with and expression involving $\cos2\theta$ is not necessarily a good idea; then you have to deal with cosines of two different angles.

A better approach is to realize that what we have is a quadratic equation: let us define $y$ to be $y=\cos\theta$. Then we can rewrite the equation as $$y^2 + y = 2$$ or $y^2 + y - 2 = 0$. We know how to solve quadratic equations: the solutions are $$\begin{align*} y_1 &= \frac{-1+\sqrt{1+8}}{2} = \frac{-1+3}{2} = 1\\ y_2 &= \frac{-1-\sqrt{1+8}}{2} = \frac{-1-3}{2} = -2. \end{align*}$$ However, now we remember that $y$ is actually $\cos\theta$, so now we want to find the solutions to $\cos\theta = 1$ and of $\cos\theta=-2$.

Since $-1\leq\cos\theta\leq 1$, the latter equation has no solutions.

So the answer is that the solutions are exactly the $\theta$ for which $\cos(\theta)=1$.

(Which we could have figured out cleverly by making the observation made by David Mitra in comments, but I wanted to give you an idea of how to approach this kind of equation if the answer is not so obvious.)

share|cite|improve this answer
Right I understand the substitution of $\cos\theta$ to $y$. I appreciate your answer and @David Mitra 's help. Is there anyway you can help me solve this without the use of $y$ instead using $\cos$ – Austin Broussard Jul 14 '12 at 23:59
@AustinBroussard: What do you mean? Using $y$ is just a device to make things easier. You can apply the quadratic formula directly to $\cos(\theta)$ to conclude that $\cos\theta=1$ is the only way this equation can be true. Then you just need the values of $\theta$ for which the cosine is $1$. If you mean you want to solve it without invoking the quadratic formula at all, then David Mitra's solution is the way to go: since $-1\leq\cos\theta\leq 1$ and $0\leq \cos^2\theta\leq 1$, the only way for $\cos^2\theta+\cos\theta$ to equal $2$ is if $\cos^2\theta=1$ and $\cos\theta=1$. – Arturo Magidin Jul 15 '12 at 0:01
I understand now what you mean. – Austin Broussard Jul 15 '12 at 0:04
@Joel: Yes. Thank you. – Arturo Magidin Jul 15 '12 at 0:36

Hint: $\cos x$ is bounded in absolute value by 1; so the equality holds only when $\cos x=1$.

Alternatively, you can think of your equation as a quadratic equation in the variable $\cos \theta$. You will see that there is only one solution in the interval $[0, 2\pi)$ (there are actually infinitely many solutions...) after you solve the quadratic (you'll obtain the equations $\cos\theta=1$ or $\cos\theta=-2$).

share|cite|improve this answer
So where would I go from $\dfrac{1+\cos^2\theta−\sin^2\theta}{2}+\cos\theta=2$ – Austin Broussard Jul 14 '12 at 23:56
@AustinBroussard I would suggest not using that approach :) – David Mitra Jul 14 '12 at 23:59

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.