# 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 $$\in$$ I

### Similar Questions

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

Solve : cos3x + sin2x – sin4x = 0

Solve : 6 – 10cosx = 3$$sin^2x$$

Find general solution of (2sinx – cosx)(1 + cosx) = $$sin^2x$$