If #sin^4theta+cos^4theta=1#, find #theta#?

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Answer 1 · 2017-09-08T04:42:17

#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#

Answer 2 · 2017-09-08T04:51:05

#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.