General solution of $(\sqrt3 - 1)\sin\theta + (\sqrt3 + 1)\cos\theta =2 $ Find general solution of the equation $(\sqrt3 - 1)\sin\theta + (\sqrt3 + 1)\cos\theta =2 $.
My approach:
Squared on both sides, formed a quadratic equation in $\cos\theta$ and finally got two solutions for theta,
$$\theta = 2n\pi \pm \frac{\pi}{6}$$
$$\theta = 2n\pi \pm \frac{\pi}{3}$$
But the answer given in my book is $$2n\pi \pm \frac{\pi}{4} + \frac{\pi}{12}$$
Pretty strange
can anyone help me?

\begin{align}
(\sqrt3 - 1)\sin\theta + (\sqrt3 + 1)\cos\theta &=2\\
\left(\frac{\sqrt3}{2} -\frac{1}{2}\right)\sin\theta+\left(\frac{\sqrt3}{2} +\frac{1}{2}\right)\cos\theta&=1\\
\left(\cos30 -\sin(30)\right)\sin\theta+\left(\cos30 +\sin(30)\right)\cos\theta&=1\\
\sin\theta\cos30 -\sin\theta\sin(30)+\cos\theta\cos30 +\cos\theta\sin(30)&=1\\
\sin(\theta+30) +\cos(\theta+30) &=1\\
\sin(\theta+\pi/6) +\cos(\theta+\pi/6) &=1\\
\text{Clearly the solution set includes $\pi/3 + 2\pi k$ and $-\pi/6+2\pi k$}\\
\text{The trick now to simplify is to take the average value}\\ (\pi/3 + -\pi/6)/2&= \pi/12\\
\text{The two solutions show up every $2\pi k$}\\
\text{The distance for either solution from $\pi/12$ is $\pi/4$ }\\
\theta &= \pi/12 + 2\pi k \pm \pi/4
\end{align}
