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I just have a quick question. It's simple but I am having trouble with it.

Solve the equation $\:\:12 \cos^2\theta - 6= \sin\theta\:\:$ for $\theta$ in $(-2\pi,2\pi)$.

I am unsure what to do with this except rearrange the terms.

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closed as off-topic by Carl Mummert, Claude Leibovici, user91500, Daniel W. Farlow, Zain Patel Jul 11 '16 at 15:55

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    $\begingroup$ Let $x=\sin \theta$. Then the equation is $12(1-x^2)-6=x$. Solve this quadratic equation. $\endgroup$ – Dietrich Burde Jul 10 '16 at 18:15
  • $\begingroup$ You must re-write $\cos^2\theta$ in terms of $\sin^2\theta$ and then you have a quadratic in form equation in $\sin\theta$. $\endgroup$ – John Wayland Bales Jul 10 '16 at 18:16
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You can change it into a quadratic in terms of $\sin\theta$ and then use the quadratic formula.

\begin{align} 12\cos^2\theta - 6 &= \sin\theta \\12(1-\sin^2\theta) - 6 &= \sin\theta \\12-12\sin^2\theta - 6 &= \sin\theta \\12\sin^2\theta +\sin\theta - 6 &= 0 \end{align}

\begin{align} \implies\quad \theta = \arcsin&\left(\frac{-1 \pm \sqrt{1+4(12)(6)}}{2(12)}\right) \\\theta = \arcsin\left(-\frac{3}{4}\right) \quad&\text{or}\quad \theta = \arcsin\left(\frac{2}{3}\right) \end{align}

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  • $\begingroup$ Square root of 289 is 17? $\endgroup$ – user305860 Jul 10 '16 at 19:00
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hint: $\cos^2 \theta = 1 -\sin^2 \theta$

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  • $\begingroup$ This isn't just a "hint"; this is the first step. $\endgroup$ – Michael Hardy Jul 10 '16 at 18:17
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Deepsea's hint:

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

$$12(1 -\sin^2 \theta)-6=\sin \theta$$

Let $x:=\sin \theta$

$$12-12x^2-6=x\\ 12x^2+x-6=0\\ \vdots$$

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