Question

How do you use the inverse functions where needed to find all solutions of the equation `2sin^2x-7sinx+3=0` in the interval `[0,2pi)`?

Answer 1

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

Explanation:

Solve this quadratic equation for sin x by using the improved quadratic formula (Socratic Search):
`2sin^2 x - 7sin x + 3 = 0`
`D = d^2 = b^2 - 4ac = 49 - 24 = 25` --> `d = +- 5`
There are 2 real roots:
`sin x = -b/(2a) +- d/(2a) = 7/4 +- 5/4`
a. sin x = 12/4 = 3 (rejected because > 1)
b. `sin x = (7 - 5)/4 = 2/4 = 1/2`
Trig table and unit circle give 2 solutions for `(0, 2pi)`
`x = pi/6`, and `x = pi - pi/6 = (5pi)/6`