Double Angle Identities question. Please help?!

The following for x in #0<=x<=180#:

#8sin^2xcos^2x=1#

1 Answer
Nghi N
Mar 30, 2018

#pi/8; (3pi)/8#

Explanation:

Apply the trig identity:
#sin 2x = 2sin x.cos x# -->
#sin^2 2x = 4sin^2 x.cos^2 x#.
The given equation becomes:
#2sin^2 2x = 1#
#sin^2 2x = 1/2#
#sin 2x = +- 1/sqrt2 = +- sqrt2/2#
a. #sin 2x = sqrt2/2#
Trig table and unit circle give 2 solutions:
1. #2x = pi/4# --> #x = pi/8#
2. #2x = (3pi)/4# --> #x = (3pi)/8#
b. #sin 2x = - sqrt2/2#.
#x = - pi/4# and #x = - (3pi)/4#
Both answers are rejected because out of interval (0, 180)
The 2 answers in this interval are: #pi/8, (3pi)/8#

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