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

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.
