# Solve $\sqrt{3}\cos2\theta+\sin2\theta-1=0$

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 $R\sin(2\theta+\alpha)$?

• Why do you wish to use these identities? Can't you start solving in $\alpha$ by stating $\alpha=2\theta$? – Martigan Nov 3 '14 at 9:34

Hint: $\sqrt{3}\cos 2\theta + \sin 2\theta = 2\sin(2\theta+\pi/3)$

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.

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.

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}

• Also, the formula is clearly incorrect - what happens when $a>1$? – nbubis Nov 3 '14 at 9:54