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How do you solve #2sin^2x + sinx =0# from #[0,2pi]#?

1 Answer

Nghi N. · AJ Speller

Jul 25, 2015

Solve #2sin^2 x + sin x = 0# # [0, 2pi]#

Explanation:

#sin x(2sin x + 1) = 0#
Use trig table and unit circle -->

a. sin x = 0 => x = 0, and #x = pi# and #x = 2pi#

b. 2sin x + 1 = 0 => #sin x = -1/2# => #x = (7pi)/6# and #x = (11pi)/6#.

Finally, within interval #[0, 2pi],# there are 5 answers:
#0, pi, 2pi, (7pi)/6 and (11pi)/6#

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