How do I find ALL solutions to the following equation?

#sin^2xcos^2x=(2-sqrt2)/16#

1 Answer
Apr 9, 2018

#x = +- 37^@08#
#x = +- 52^@92#

Explanation:

Use trig identity: sin 2x = 2sin x.cos x
#sin^2 2x = 4sin^2 x.cos ^2 x#
#sin^2 x.cos^2 x = (sin^2 2x)/4 = (2 - sqrt2)/16#
#sin^2 2x = (4(2 - sqrt2))/32 = (2 - sqrt2)/8#
#sin 2x = +- sqrt(2 - sqrt2)/(2sqrt2)#
Calculator gives:
#2x = +- 0.77/2.82 = +- 0.273 #
Calculator and unit circle give 4 solutions:
a. 2x = 0.273 --> #2x = +- 74^@15# --># x = +- 37^@08#
b. 2x = - 0.273 --> #2x = +- 105.84# --> #x = +- 52^@92#