Solve : $$\sqrt{3} cos \theta$$ + $$sin \theta$$ = $$\sqrt{2}$$

Solution :

We have,

$$\sqrt{3} cos \theta$$ + $$sin \theta$$ = $$\sqrt{2}$$            ………….(i)

This is of the form $$a cos\theta$$ + $$b sin \theta$$ = c, where a = $$\sqrt{3}$$, b = 1 and c = $$\sqrt{2}$$.

Let a = $$r cos\alpha$$ and b = $$r sin\alpha$$. Then,

$$\sqrt{3}$$ = $$r cos\alpha$$  and  1 = $$r sin\alpha$$

$$\implies$$  r = $$\sqrt{a^2 + b^2}$$ = $$\sqrt{(\sqrt{3})^2 + 1^2}$$ = 2 and $$tan \alpha$$ = $$1\over \sqrt{3}$$ $$\implies$$  $$\alpha$$ = $$\pi\over 6$$

Substituting a = $$\sqrt{3}$$ = $$r cos\alpha$$  and  b = 1 = $$r sin \alpha$$ in the equation (i) it reduces to

$$r cos\alpha cos\theta$$ + $$r sin\alpha sin\theta$$ = $$\sqrt{2}$$

$$\implies$$ $$r cos(\theta – \alpha)$$ = $$\sqrt{2}$$

$$\implies$$  $$2 cos(\theta – {\pi\over 6}$$ = $$\sqrt{2}$$

$$\implies$$  $$cos(\theta – {\pi\over 6})$$ = $$1\over \sqrt{2}$$

$$\implies$$  $$cos(\theta – {\pi\over 6})$$  = $$cos{\pi\over 4}$$

$$\implies$$  $$\theta – {\pi\over 6}$$ = $$2n\pi \pm {\pi\over 4}$$,  n $$\in$$  Z.

$$\implies$$  $$\theta$$ = $$2n\pi \pm {\pi\over 4} + {\pi\over 6}$$,  n $$\in$$  Z.

$$\implies$$  $$\theta$$ = $$2n\pi + {\pi\over 4} + {\pi\over 6}$$  or,  $$\theta$$ = $$2n\pi – {\pi\over 4} + {\pi\over 6}$$

$$\implies$$  $$\theta$$ = $$2n\pi + {5\pi\over 12}$$  or  $$2n\pi – {\pi\over 12}$$

Hence,  $$\theta$$ = $$2n\pi + {5\pi\over 12}$$  or,  $$\theta$$ = $$2n\pi – {5\pi\over 12}$$,  where n $$\in$$  Z.