# Trigonometric Equations Questions

## Find the number of solutions of tanx + secx = 2cosx in [0, $$2\pi$$].

Solution : Here, tanx + secx = 2cosx       $$\implies$$     sinx + 1 = $$2cos^2x$$ $$\implies$$ $$2sin^2x$$ + sinx – 1 = 0     $$\implies$$    sinx = $$1\over 2$$, -1 But sinx = -1 $$\implies$$ x = $$3\pi\over 2$$ for which tanx + secx = 2cosx  is not defined. Thus, sinx = …

## If $${1\over 6}sin\theta$$, $$cos\theta$$ and $$tan\theta$$ are in G.P. then the general solution for $$\theta$$ is

Solution : Since, $${1\over 6}sin\theta$$, $$cos\theta$$ and $$tan\theta$$ are in G.P. $$\implies$$ $$cos^2\theta$$ = $${1\over 6}sin\theta$$.$$cos\theta$$ $$\implies$$ $$6cos^3\theta$$ + $$cos^2\theta$$ – 1 = 0 $$\therefore$$   ($$2cos\theta – 1$$)($$3cos^2\theta$$ + $$2cos\theta$$ + 1) = 0 $$cos\theta$$ = $$1\over 2$$       (other values are imaginary) $$cos\theta$$ = $$cos\pi\over 3$$   $$\theta$$ = $$2n\pi \pm {\pi\over 3}$$,  n …

## Solve : cos3x + sin2x – sin4x = 0

Solution : we have, cos3x + (sin2x – sin4x) = 0 = cos3x – 2sinx.cos3x = 0 $$\implies$$  (cos3x)(1 – 2sinx) = 0 $$\implies$$  cos3x = 0  or  sinx = $$1\over 2$$ $$\implies$$  cos3x = 0 = cos$$\pi\over 2$$  or  sinx = $$1\over 2$$ = sin$$\pi\over 6$$ $$\implies$$  3x = 2n$$\pi$$ $$\pm$$ $$\pi\over 2$$  or  …

Solution : we have, 6 – 10cosx = 3$$sin^2x$$ $$\therefore$$  6 – 10cosx = 3 – 3$$cos^2x$$ $$\implies$$  3$$cos^2x$$ – 10cosx + 3 = 0 $$\implies$$  (3cosx-1)(cosx-3) = 0  $$\implies$$  cosx = $$1\over 3$$ or cosx = 3 Since cosx = 3 is not possible as -1 $$\le$$ cosx $$\le$$ 1 $$\therefore$$  cosx = $$1\over … ## Find general solution of (2sinx – cosx)(1 + cosx) = \(sin^2x$$

Solution : (2sinx – cosx)(1 + cosx) – (1 – $$cos^2x$$) = 0 $$\therefore$$ (1 + cosx)(2sinx – cosx – 1 + cosx) = 0 $$\therefore$$  (1 + cosx)(2sinx – 1) = 0 $$\implies$$ cosx = -1  or  sinx = $$1\over 2$$ $$\implies$$  cosx = -1 = cos$$\pi$$  $$\implies$$  x = 2n$$\pi$$ + $$\pi$$ = …