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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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    @Alex The identity should be $\cos^2 \theta + \sin^2 \theta = 1$, with a plus.2012-03-10

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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    @MichaelHardy. Ops, I'll edit it.2012-03-10