# Trigonometric Equations Questions

Solution : The general solution of $$sin \theta$$ = $$sin \alpha$$ is given by $$\theta$$ = $$n\pi + (-1)^n \alpha$$,  n $$\in$$ Z. Proof : We have,  $$sin \theta$$ = $$sin \alpha$$ $$\implies$$  $$sin \theta$$ – $$sin \alpha$$ = 0 $$\implies$$   $$2 sin ({\theta – \alpha\over 2}) cos({\theta + \alpha\over 2})$$ = 0 $$\implies$$  $$sin … ## 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, $$\sqrt{3}$$ = …

## What is the General Solution of $$sin^2 \theta$$ =$$sin^2 \alpha$$ ?

Solution : The general solution of $$sin^2 \theta$$ = $$sin^2 \alpha$$ is given by $$\theta$$ = $$n\pi \pm \alpha$$, n $$\in$$ Z. Proof : We have, $$sin^2 \theta$$ =$$sin^2 \alpha$$ $$\implies$$  $$2 sin^2 \theta$$ =$$2 sin^2 \alpha$$ $$\implies$$  $$1 – cos 2\theta$$ = $$1 – cos 2\alpha$$ $$\implies$$   $$cos 2\theta$$ = $$cos 2\alpha$$ $$\implies$$  $$2\theta$$ …

## What is the General Solution of $$cos^2 \theta$$ =$$cos^2 \alpha$$ ?

Solution : The general solution of $$cos^2 \theta$$ = $$cos^2 \alpha$$ is given by $$\theta$$ = $$n\pi \pm \alpha$$, n $$\in$$ Z. Proof : We have, $$cos^2 \theta$$ =$$cos^2 \alpha$$ $$\implies$$  $$2 cos^2 \theta$$ =$$2 cos^2 \alpha$$ $$\implies$$  $$1 + cos 2\theta$$ = $$1 + cos 2\alpha$$ $$\implies$$   $$cos 2\theta$$ = $$cos 2\alpha$$ $$\implies$$  $$2\theta$$ …

Solution : The general solution of $$tan^2 \theta$$ = $$tan^2 \alpha$$ is given by $$\theta$$ = $$n\pi \pm \alpha$$, n $$\in$$ Z. Proof : We have, $$tan^2 \theta$$ =$$tan^2 \alpha$$ $$\implies$$  $$1 – tan^2\theta\over 1 + tan^2 \theta$$ =$$1 – tan^2\alpha\over 1 + tan^2 \alpha$$ $$\implies$$   $$cos 2\theta$$ = $$cos 2\alpha$$ $$\implies$$  $$2\theta$$ = $$2n\pi … ## What is the General Solution of \(Cot \theta$$ = 0 ?

Solution : The general solution of $$cot \theta$$ = 0 is given by $$\theta$$ = $$(2n + 1){\pi\over 2}$$, n $$\in$$ Z. Proof : We have, $$cot \theta$$ = $$OM\over PM$$ $$\therefore$$   $$cot \theta$$ = 0 $$\implies$$  $$OM\over PM$$ = 0 $$\implies$$ OM = 0 $$\implies$$  OP coincides with OY or OY’ $$\implies$$  $$\theta$$ = …

## What is the General Solution of $$Cos \theta$$ = 0 ?

Solution : The general solution of $$cos \theta$$ = 0 is given by $$\theta$$ = $$(2n + 1){\pi\over 2}$$, n $$\in$$ Z. Proof : We have, $$cos \theta$$ = $$PM\over OP$$ $$\therefore$$   $$cos \theta$$ = 0 $$\implies$$  $$OM\over OP$$ = 0 $$\implies$$ OM = 0 $$\implies$$  OP coincides with OY or OY’ $$\implies$$  $$\theta$$ =  …

## What is the General Solution of $$Tan \theta$$ = 0 ?

Solution : The general solution of $$tan \theta$$ = 0 is given by $$\theta$$ = $$n\pi$$, n $$\in$$ Z. Proof : We have, $$tan \theta$$ = $$PM\over OM$$ $$\therefore$$   $$tan \theta$$ = 0 $$\implies$$  $$PM\over OM$$ = 0 $$\implies$$ PM = 0 $$\implies$$  OP coincides with OX or OX’ $$\implies$$  $$\theta$$ = 0, $$\pi$$, $$2\pi$$, …