Question

How do you find the exact solutions of the equation `sin4x=-2sin2x` in the interval `[0,2pi)`?

Answer 1

`0, pi/2, pi, (3pi)/2`

Explanation:

Bring the equation to standard form:
sin 4x + 2sin 2x = 0
Substitute (sin 4x) by (2sin 2x.cos 2x) (trig identity):
2sin 2x.cos 2x + 2sin 2x = 0
2sin 2x(cos 2x + 1) = 0
a. sin 2x = 0 -->
Unit circle gives 3 solutions for 2x -->
2x = 0, --> `x = 0`
2x = pi, `x = pi/2` -->
2x = 2pi --> `x = pi`
b. cos 2x = - 1
Unit circle give 2 solutions for 2x -->
`2x = pi` --> `x = pi/2`
`2x = 3pi` --> `x = (3pi)/2`
Answers for `(0, 2pi)`
`0, pi/2, pi, (3pi)/2`
General answers: `x = k(pi/2)`