<s>[INST] 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 $R\sin(2\theta+\alpha)$?
 [/INST] 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.
</s>