Question

How do you solve `3sin^2(t)+ 10sin(t)+ 2 = 0`?

Answer 1

`192^@34, 347^@66`

Explanation:

Solve this quadratic equation for sin t.
`3sin^2 t + 10 sin t + 2 = 0`
`D = b^2 - 4ac = 100 - 24 = 76` --> `d = +- 2sqrt19`
There are 2 real roots:
`sin t = - b/(2a )+- d/(2a) = - 10/6 +- (2sqrt19)/6 = (-5 +- sqrt19)/3`
The root `sin t = (-5 - sqrt19)/3 = - 3.12` is rejected as < -1.
The root `sin t = (-5 + sqrt19)/3 = - 0.213` .
Calculator and the unit circle give 2 solution arcs:
`t1 = -12^@34` , or `t1 = 347^@66` (co-terminal), and
`t2 = 180 - (-12.34) = 180 + 12.34 = 192^@34`

Answers for `(0, 2pi):`
`192^@34 and 347^@66`
For general answers, add `k360^@`