How do you find the general solutions for #4sin^3x=2sin^2x-2sinx-1=0#?

1 Answer
Nghi N.
Aug 22, 2015

Solve: f(x) = 4sin^3x + 2sin ^2 x - 2sin x - 1 = 0

Explanation:

Call sin x = t -->
#4t^3 + 2t^2 - 2t - 1 = 0#
#2t^2 (2t + 1) - (2t + 1) = 0#
#(2t + 1) (2t^2 - 1) = (2t + 1)(sqrt2t - 1)(sqrt2t + 1) = 0#

a. #t = sin x = - 1/2# --> #x = -pi/6# and #x = -(5pi)/6#
b. #t = sin x = sqrt2/2# --> #x = pi/4# and #x = (3pi)/4#
c #t = sin x = -sqrt2/2# --> #x = -pi/4# and #x = - (3pi)/4#
Within the interval (0, 2pi), there are 6 answers:
#x = (7pi)/6 + 2kpi#
#x = (11pi)/6 + 2kpi#
#x = (pi)/4 + 2kpi#
#x = (3pi)/4 + 2kpi#
#x = (5pi)/4 + 2kpi#
#x = (7pi)/4 + 2kpi#

Related questions
See all questions in Solving Trigonometric Equations
Impact of this question
9433 views around the world
You can reuse this answer
Creative Commons License