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I tried using the identities $\cos2\theta=1-2\sin^2\theta$ and $\sin2\theta=2\sin\theta\cos\theta$. These give

$\sqrt{3}(1-2\sin^2\theta)+2\sin\theta\cos\theta-1=0$

which doesn't seem to lead anywhere. Perhaps I must equate the function to something like $R\sin(2\theta+\alpha)$?

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    $\begingroup$ Why do you wish to use these identities? Can't you start solving in $\alpha$ by stating $\alpha=2\theta$? $\endgroup$ – Martigan Nov 3 '14 at 9:34
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Hint: $\sqrt{3}\cos 2\theta + \sin 2\theta = 2\sin(2\theta+\pi/3)$

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It's easier if you only use one identity: $$\sqrt{3}\sqrt{1-\sin^2 2\theta}+\sin 2\theta=1$$ Write $t=\sin 2\theta$ to get: $$3(1-t^2)=(1-t)^2$$ Which is just a quadratic equation.

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When you have an equation of the form $$ a\sin^2\theta+b\sin\theta\cos\theta+c\cos^2\theta+d=0 $$ you can use $1=\cos^2\theta+\sin^2\theta$ and write the equation as $$ (a+d)\sin^2\theta+b\sin\theta\cos\theta+(c+d)\cos^2\theta=0 $$ If $a+d=0$, this factors; otherwise $\cos\theta=0$ is not a solution and so you can transform it into $$ (a+d)\tan^2\theta+b\tan\theta+(c+d)=0 $$ that's quadratic.

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In general you can write: \begin{align} a\cos\theta+b\cos\theta&= \sqrt{a^2+b^2}\left( \frac{a}{\sqrt{a^2+b^2}}\cos\theta+\frac{b}{\sqrt{a^2+b^2}}\sin\theta \right)\\[2ex] &=\sqrt{a^2+b^2}(\sin\alpha\cos\theta+\cos\alpha\sin\theta) \end{align}

Hope this is helpful.

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  • $\begingroup$ Panda, the whole point of LaTeX formatting is so that you do not need to copy and paste images, and moreover answers are then searchable. $\endgroup$ – nbubis Nov 3 '14 at 9:53
  • $\begingroup$ Also, the formula is clearly incorrect - what happens when $a>1$? $\endgroup$ – nbubis Nov 3 '14 at 9:54
  • $\begingroup$ I added the missing factor (you seemed to have forgotten it) and translated into MathJax. $\endgroup$ – egreg Nov 3 '14 at 9:58
  • $\begingroup$ @nbubis:suppose a=10000,b=1 which sentence is wrong a/(...) is always under 1 so this is always true. $\endgroup$ – Panda Nov 3 '14 at 16:18
  • $\begingroup$ @Panda - with egreg's correction, this is now correct. $\endgroup$ – nbubis Nov 3 '14 at 16:25

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