Solve #sin^2x + sin^2 2x = 1# ?

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Answer 1 · 2018-04-28T21:39:56

#pi/2+pin#
#pi/6+pin#
#(5pi)/6+pin#

Where n is an element of all integers

#sin^2x + sin^2 2x = 1#

#sin^2x+sin2x*sin2x=1#

Using sine double angle identity:
#sin^2x+(2sinxcosx)(2sinxcosx)=1#

#sin^2x+4sin^2xcos^2x=1#

Using pythagorean identity:
#sin^2x+4sin^2x(1-sin^2x)=1#

#sin^2x+4sin^2x-4sin^4x=1#

#-4sin^4x+5sin^2x-1=0#

#-(4sin^4x-5sin^2x+1)=0#

#-(4sin^4x-4sin^2x-sin^2x+1)=0#

#-(4sin^2x(sin^2x-1)-1(sin^2x-1))=0#

#-(4sin^2x-1)(sin^2x-1)=0#

#sinx=+-1#
#x=pi/2, (3pi)/2#

#sinx= +-1/2#
#x= pi/6, (5pi)/6, (7pi)/6, (11pi)/6#

graph{(sinx)^2+(sin(2x))^2-1 [-10, 10, -5, 5]}