Solve: 2(sin^4x + cos^4x) = 1

Question

with intervals of: [-pi ≤ x ≤ pi ]

Answer 1

2(sin^4x + cos^4x ) = 1

2(( sin^2x)^2 + cos^2x ) = 1

2( ( 1 - cos^2x )^2 + cos^4x) = 1

2(( 1 -2cos^2x + cos^4x ) + cos^4x ) = 1

2( 1 - 2cos^2x + 2 cos^4x ) - 1 = 0

2 - 4cos^2x + 4cos^4x - 1 = 0

1-4cos^2x + 4cos^4x = 0

( cos^2x = t )

1 - 4t + 4t^2 = 0

( 1 - 2 t )^2 = 0

1 - 2t = 0

2t = 1

t = 1/2

cos^2x =  1/2

cosx = 1/sqrt2

cosx =  sqrt2/2 ⇒

x = 45 degree,  becouse cos45 = sqrt2/2

Answer 2

2sin^4(x) +2cos^4(x)+sin^2(2x)

Answer 3

pi/3 and 2pi/3