Question

How do solve the equation, 3 sinx Sin2x = 2cos³x, giving all solutions in the interval 0<x<2pi, in terms of pi?

Answer 1

`pi/6; (5pi)/6; (7pi)/6; (11pi)/6`

Explanation:

Replace in the equation sin 2x by (2sin x.cos x) -->
`3sin x(2sin x.cos x) = 2cos ^3 x`
Divide both sides by 2cos x (condition `cos x != 0`):
`3sin^2 x = cos^2 x`
Divide both sides by `cos^2 x`
`3tan^2 x = 1` --> `tan^2x = 1/3`
`tan x = +- sqrt3/3`
Trig table and unit circle give 4 solutions for (0. 2pi)
a. `tan x = sqrt3/3` --> `x = pi/6` and `x = pi/6 + pi = (7pi)/6`
b. `tan x = - sqrt3/3` --> `x = (5pi)/ 6` , and
`x = (5pi)/6 + pi = (11pi)/6`