How do you solve #sin^2 x - cos^2 x=0# for x in the interval [0,2pi)?

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Answer 1 · 2016-03-10T23:20:27

#x in {pi/4, (3pi)/4, (5pi)/4, (7pi)/4}#

Applying the identity #cos^2(x)-sin^2(x) = cos(2x)# we have

#sin^2(x)-cos^2(x) = -cos(2x)#

In general,

#cos(u) = 0 <=> u = (npi)/2# for some #n in ZZ#

Thus we have

#sin^2(x) - cos^2(x) = 0#

#=>-cos(2x) = 0#

#=> 2x = (npi)/2# for #n in ZZ#

#=> x = (npi)/4# for #n in ZZ#

Restricting our values to the interval #[0, 2pi]# gives our final result:

#x in {pi/4, (3pi)/4, (5pi)/4, (7pi)/4}#