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I have the following equation, and I have been stumped on it for a long time now, I was wondering if I could get some hints in attempting to solve it.

$$ 2\cos^2\theta-\cos\theta-1 = \sin^2\theta $$

Solved!

Use the hyperbolic function (Thanks svenkatr):

$$ \cos^2\theta - \sin^2\theta = 1 $$

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    $\begingroup$ Add $\cos^2(\theta)$ to both sides, and reduce it to quadratic equation in $\cos(\theta)$. $\endgroup$
    – Sasha
    Mar 9, 2012 at 22:47
  • $\begingroup$ @Sasha: Please read this for reasons why it might be better to add an answer rather than commenting: meta.math.stackexchange.com/questions/1559/… $\endgroup$
    – Aryabhata
    Mar 9, 2012 at 22:50
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    $\begingroup$ You have a trigonometric equation, which you want to solve. An identity is an equality that holds for every possible value of the variable(s). $\endgroup$ Mar 9, 2012 at 23:03
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    $\begingroup$ @Alex - $\cos^2 \theta + \sin^2 \theta = 1$. We are dealing with trigonometric, not hyperbolic functions. $\endgroup$
    – svenkatr
    Mar 9, 2012 at 23:28
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    $\begingroup$ @Alex The identity should be $\cos^2 \theta + \sin^2 \theta = 1$, with a plus. $\endgroup$
    – TMM
    Mar 10, 2012 at 1:37

2 Answers 2

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Write $\sin^2 \theta = 1- \cos^2 \theta$, simplify to get a quadratic equation in $\cos \theta$ and solve the equation.

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As Sasha pointed out, one can simply add $\cos^2\theta$ to both sides and solve for the quadratic equation. Here is what occurs: $$2\cos^2\theta-\cos\theta-1+\cos^2\theta=\sin^2\theta+\cos^2\theta$$ $$3\cos^2\theta-\cos\theta-1=1$$ $$3\cos^2\theta-\cos\theta-2=0$$ Let $w=\cos\theta$. Then we get: $$3w^2-w-2=0$$ $$3w^2-3w+2w-2=0$$ $$3w(w-1)+2(w-1)=0$$ $$(3w+2)(w-1)=0$$ So $w=\frac{-2}{3}$ or $w=1$, which means $\cos\theta=\frac{-2}{3}$ or $\cos\theta=1$. This gives us the answers $\theta=2k\pi, k\in\mathbb{Z}$ and $\theta=\pm\cos^{-1}\left(\frac{-2}{3}\right)+2k\pi, k\in \mathbb{Z}$

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  • $\begingroup$ Not quite a complete answer: it omitted $-\cos^{-1}(-2/3)$, and also omitted to say $\cos^{-1}(-2/3)+2k\pi$ as in the case of $\theta=0$. $\endgroup$ Mar 10, 2012 at 0:57
  • $\begingroup$ @MichaelHardy. Ops, I'll edit it. $\endgroup$
    – E.O.
    Mar 10, 2012 at 1:18

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