If 2#cos^2x# - 2#sin^2x# = 1 , what is the value of #x#?

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Answer 1 · 2016-03-07T05:06:41

#x = +-pi/6 + npi# for #n in ZZ#

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

#1 = 2cos^2(x)-2sin^2(x)#

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

#=2cos(2x)#

#=> cos(2x) = 1/2#

#=> 2x = +-pi/3+2npi# for #n in ZZ#

#:. x = +-pi/6 + npi# for #n in ZZ#