How do you solve #4sin^2x-1=0# in the interval #0<=x<=2pi#?

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How do you solve #4sin^2x-1=0# in the interval #0<=x<=2pi#?

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Answer 1 · 2016-09-12T13:01:00

#pi/6, (5pi)/6, (7pi)/6 and (11pi)/6#

Explanation:

#4sin^2 x - 1 = 0#
#sin^2 x = 1/4#
#sin x = +- 1/2#
Use trig table of special arcs, and unit circle -->
a. When #sin x = 1/2# --> #x = pi/6 and x = (5pi)/6#
b. When #sin x = - 1/2# --> #x = (7pi)/6 and x = (11pi)/6#

Answers for #(0, 2pi)#
#pi/6, (5pi)/6, (7pi)/6, (11pi)/6#

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How do you solve #4sin^2x-1=0# in the interval #0<=x<=2pi#?

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