Question

How do you find all the solutions for `2 (sin^2)2x + 5 sin(2x) - 3 = 0`?

Answer 1

`x = pi/12 + kpi`

Explanation:

Call sin 2x = T, and solve this quadratic equation for T:
`2T^2 + 5T - 3 = 0`
`D = d^2 = b^2 - 4ac = 25 + 24 = 49` --> `d = +- 7`
There are 2 real roots:
`T = -b/(2a) +- d/(2a) = - 5/4 +- 7/4 = (-5 +- 7)/4`
`T1 = 2/4 = 1/2` and `T2 = - 12/4 = -3`
Next, use trig table and unit circle:
a. `sin 2x = T1 = 1/2`
`2x = pi/6 + 2kpi`
`x = pi/12 + kpi`
b. sin 2x = T2 = -3
Solution rejected since < - 1