How do you solve #sin^2x < 1/2#?

Solution:

2 Answers
Douglas K. · sara
May 31, 2018

Given: #sin^2(x)<1/2#

When we take the square root of both sides, we obtain two equations:

#sin(x)> -1/sqrt2# and #sin(x) < 1/sqrt2#

Take the inverse sine of both sides of both equations:

#x> sin^-1(-1/sqrt2) # and #x < sin^-1(1/sqrt2)#

We know that the primary values are #-pi/4 = sin^-1(-1/sqrt2)# and #pi/4 = sin^-1(1/sqrt2)#:

#x> -pi/4# and #x < pi/4#

This can be written as a single interval:

#-pi/4 < x < pi/4#

We know that it is periodic with integer multiples of #pi#:

#-pi/4 + npi < x < pi/4 + npi, n in ZZ#

Nghi N.
Jun 1, 2018

#(-pi/4, pi/4), and ((3pi)/4 , (5pi)/4)#

Explanation:

First, find the 4 end-points, by solving #f(x) = sin^2 x = 1/2#
a. #sin x = sqrt2/2# --> #x = pi/4 and x = (3pi)/4#
b. #sin x = - sqrt2/2# --> #x = - pi/4, and x = (5pi)/4#
Now, solve the trig inequality by using the unit circle as proof.
#f(x) = sin^2 x < 1/2#
#Isin xI < sqrt2/2#
On the unit circle, #Isin xI < sqrt2/2#, when the arc x varies inside the
2 open intervals, that are the answers:
#(-pi/4, pi/4) and ((3pi)/4, (5pi)/4)#

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