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

2 Answers
Sep 8, 2017

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

Sep 8, 2017

x = (npi)/2, where n is an integer.

Explanation:

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.