How do you solve #2sin^2(x)+3cos(x)=0# on the interval [0,2pi]?

1 Answer
Mar 20, 2016

#(2pi)/3, (4pi/3)#

Explanation:

Replace in the equation #sin^2 x# by #(1 - cos^2 x) #-->
#2(1 - cos^2 x) + 3cos x = 0#
#-2cos^2 x + 3cos x + 2 = 0#
Solve this equation for cos x.
#D = d^2 = b^2 - 4ac = 9 + 16 = 25# --> #d = +- 5#
#cos x = -b/(2a) +- d/(2a) = 3/4 +- 5/-4 = 3/4 +- 5/4#
a. #cos x = 8/4 = 2# (rejected because > 1)
b. #cos x = -2/4 = -1/2#
#cos x = -1/2# --> #x = +- (2pi)/3#
Answers for the interval #(0, 2pi)#:
#(2pi)/3, (4pi)/3# (co-terminal to #(-2pi)/3)#