<s>[INST] 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?
 [/INST] $$2=(\tan60^\circ-\tan45^\circ)\sin\theta+(\tan60^\circ+\tan45^\circ)\cos\theta$$
$$\iff\dfrac1{\sqrt2}=\sin15^\circ\sin\theta+\sin105^\circ\cos\theta$$
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 $A=60^\circ$
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