Question

How do you solve `4sin^4x+3sin^2x-1=0` and find all solutions in the interval `0<=x<360`?

Answer 1

`pi/6; (5pi)/6; (7pi)/6; (11pi)/6`

Explanation:

Call `sin^2 x = T`, we get a quadratic equation in T. Solve this quadratic equation for T:
`4T^2 + 3T - 1 = 0`
Since a - b + c = 0, use short cut. One real root is T = - 1 and the other is `T = - c/a = 1/4`.
Solve:
a. `T = sin^2 x = - 1` , rejected as imaginary number

b. `T = sin^2t = 1/4` --> `sin t = +- 1/2`
Use Trig Table and Unit Circle:
c. `sin t = - 1/2`
`t = (7pi)/6 and t = (11pi)/6`
d. `sin t = 1/2`
`t = pi/6 and t = (5pi)/6`
Answers for `(0, 2pi)`
`pi/6; (5pi)/6; (7pi)/6; (11pi)/6`