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?
 A: \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}
A: In general, to solve $a \cos x +b\sin x=c$. Divide both sides by $\sqrt{a^2+b^2}$ and then let $\sin \alpha =\frac{a}{\sqrt{a^2+b^2}}$ to get
$$\sin (x+\alpha)=\frac{c}{\sqrt{a^2+b^2}}$$
Now solve for $x$.
A: $$2=(\tan60^\circ-\tan45^\circ)\sin\theta+(\tan60^\circ+\tan45^\circ)\cos\theta$$
$$\iff\dfrac1{\sqrt2}=\sin15^\circ\sin\theta+\sin105^\circ\cos\theta$$
As $\sin105^\circ=\sin(90^\circ+15^\circ)=\cos15^\circ,$
We have $\cos45^\circ=\cos(\theta-15^\circ)$
$\implies\theta-15^\circ=360^\circ n\pm45^\circ$ where $n$ is any integer
More generally, $$\sec A=(\tan A-\tan45^\circ)\sin\theta+(\tan A+\tan45^\circ)\cos\theta$$
$$\implies\cos\{\theta-(A-45^\circ)\}=\cos45^\circ$$
Here $A=60^\circ$
