How do you solve #sin^2x-1/2cosx+cosxsin^2x=1/2# in the interval #0<=x<=2pi#?

1 Answer
Nghi N.
Aug 19, 2016

#pi; pi/4, (3pi)/4; (5pi)/4; (7pi)/4#

Explanation:

Re- write the equation:
#sin^2 x + cos x.sin^2 x - 1/2 - (1/2)cos x = 0#
#sin^2 x(1 + cos x) - (1/2)(1 + cos x) = 0#
#(1 + cos x)(sin^2 x - 1/2) = 0#
Trig table and unit circle give 2 solutions:
#1. (1 + cos x) = 0# --> #cos x = - 1# -->
#x = pi#
2. #sin^2 x = 1/2# --> #sin x = +- 1/sqrt2 = +- sqrt2/2#
Trig table and unit circle give 4 solution arcs:
a. #sin x = sqrt2/2# --> #x = pi/4 and x = (3pi)/4#
b. #sin x = - sqrt2/2# --> #x = (5pi)/4 and x = (7pi)/4#

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