Question

How do you solve `2sin^2x-5sinx-3 = 0`?

Answer 1

`x = (7pi)/6 + 2kpi`
`x = (11pi)/6 + 2kpi`

Explanation:

`f(x) = 2sin^2 x - 5sin x - 3 = 0`
Solve this quadratic equation for sin x.
`D = d^2 = b^2 - 4ac = 25 + 24 = 49` --> `d = +- 7`
There are 2 real roots:
`sin x = -b/(2a) +- d/(2a) = 5/4 +- 7/4` -->
`sin x = - 2/4 = - 1/2`, and sin x = 3 (rejected as > 1)
`sin x = - 1/2`
Trig table and unit circle give 2 solutions for x:
`x = - pi/6 + 2kpi`, or `x = (11pi)/6 + 2kpi` (co-terminal)
and `x = pi - (-pi/6) = (7pi)/6 + 2kpi`