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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.

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    $\begingroup$ Hint: $\cos x$ is bounded in absolute value by 1; so the equality holds only when $\cos x=1$. $\endgroup$ Commented Jul 14, 2012 at 23:50
  • $\begingroup$ This equation looks like a quadratic polynomial, with $cos\theta$ as the variable... $\endgroup$ Commented Jul 14, 2012 at 23:52

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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.)

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  • $\begingroup$ 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 $\cos$ $\endgroup$ Commented Jul 14, 2012 at 23:59
  • $\begingroup$ @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 $\cos\theta=1$. $\endgroup$ Commented Jul 15, 2012 at 0:01
  • $\begingroup$ I understand now what you mean. $\endgroup$ Commented Jul 15, 2012 at 0:04
  • $\begingroup$ @Joel: Yes. Thank you. $\endgroup$ Commented Jul 15, 2012 at 0:36
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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 $\cos\theta=-2$).

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  • $\begingroup$ So where would I go from $\dfrac{1+\cos^2\theta−\sin^2\theta}{2}+\cos\theta=2$ $\endgroup$ Commented Jul 14, 2012 at 23:56
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    $\begingroup$ @AustinBroussard I would suggest not using that approach :) $\endgroup$ Commented Jul 14, 2012 at 23:59

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