Can someone give me a clue, how can i solve this? I want something like: #x=pi/4+kpi# ---> Solve this: #2sin^2(x)-sin^2(2x)=cos^2(2x)#

Question

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

It should be:

#x=pi/4+kpi# AND #x=(3pi)/4+kpi#

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Answer 1 · 2017-04-06T16:01:44

See below.

You can write also

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

so

#sin^2x=1/2->sinx=pm sqrt(2)/2#

so

#x = pm pi/4+2kpi# for #k in ZZ#

Answer 2 · 2017-04-06T16:08:21

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

Add #sin^2(2x)# to both sides:

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

Substitute 1 for #sin^2(2x)+cos^2(2x)#:

#2sin^2(x)=1#

#sin^2(x)=1/2#

#sin(x) = +-sqrt(2)/2#

#x = sin^-1(sqrt(2)/2) and x = sin^-1(-sqrt(2)/2)#

#x = pi/4+kpi and x = (3pi)/4+ kpi; k in ZZ#