the eq: sin^6 x+cos^6 x=a , has a real sol if { 1/4<=a<=1}?

Question

HOW?????

Answer 1

let sin^2(x)=t

then t^3 + (1-t)^3 = a =>
t^3+1-t^3-3t+3t^2 = a or 3t^2 - 3t +(1-a) = 0 which is a quadratic equation having real solution if 9-4(1-a)3>=0
or, 9 - 12 + 12a >0 or 12a>3 or a>1/4

further, we have to see conditions under which 0<=t<=1

Answer 2

sin^6x+c0s^6x=(sin^2x+cos^2x)(sin^4x-sin^2xcos^2x+cos^4x)
[(sin^2x+cos^2x)^2-3sin^2xcos^2x)
[1-3sin^2x(1-sin^2x)]
=(1-3sin^2x+3sin^4x)
put sin^2x=t
3t^2-3t+1
use the quadratic formula and find the condition fort real t
and revert back to sin^2x