If #sin^4theta+cos^4theta=1#, find #theta#?
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"What polyatomic ion has a positive charge?"
#sin^4theta+cos^4theta=1#
#=>(sin^2theta+cos^2theta)^2-2sin^2thetacos^2theta=1#
#=>1^2-2sin^2thetacos^2theta=1#
#=>2sin^2thetacos^2theta=0#
#=>sinthetacostheta=0#
So
when #sintheta=0#
#=>theta=npi" where " nin ZZ#
when #costheta=0#
#=>theta=((2n+1)pi)/2" where " nin ZZ#
combining we get
#theta=(kpi)/2" where " k in ZZ#
#x = (npi)/2#, where #n# is an integer.
let we consider,
#sin^2 x + cos^2 x =1#
#(sin^2 x + cos^2 x)^2 =1^2#
#sin^4 x + cos^4x + 2 sin^2 x cos^2 x = 1#
#sin^4 x + cos^4x = 1 - 2 sin^2 x cos^2 x#
since, #sin^4 x + cos^4x = 1#, then
#1 = 1 - 2 sin^2 x cos^2 x#
#2 sin^2 x cos^2 x = 0#
#1/2(4 sin^2 x cos^2 x) = 0#
#1/2(2 sin x cos x) ^2= 0#
#->(2 sin x cos x) = sin 2 x#, therefore
#1/2 (sin 2 x)^2 = 0# #->##sin^2 2x = 0#
#sin 2x = 0#
#2 x = 0, pi, 2 pi, 3pi, 4pi, ...#
#x = 0, pi/2, pi, (3pi)/2, 2pi. ...#
#x = (npi)/2#, where #n# is an integer.
