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

1 Answer
Nghi N.
Oct 11, 2016

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

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