sin^4 x + cos^4 x = (sin^2 x + cos^2 x)^2 - 2sin^2 x.cos^2 x =sin4x+cos4x=(sin2x+cos2x)2−2sin2x.cos2x=
= 1 - 2sin^2 x.cos^2 x = 1 - (sin^2 2x)/2 >= 1/2=1−2sin2x.cos2x=1−sin22x2≥12
The inequality becomes:
(sin^2 (2x))/2 <= 1/2sin2(2x)2≤12
sin^2 (2x) <= 1sin2(2x)≤1
sin 2x <= +- 1sin2x≤±1
Solve this inequality in 2 cases:
a. sin 2x <= 1sin2x≤1 -->
The answer is undefined, because sin 2x always < 1, regardless of
the value of x.
b. sin 2x <= - 1sin2x≤−1 (rejected because < - 1)
Conclusion: The inequality is always true.
Check.
x = pi/6 --> sin^4x + cos^4 x = 1/16 + 9/16 = 10/16 = 5/8 > 1/2x=π6−→sin4x+cos4x=116+916=1016=58>12
x = (2pi)/3 --> sin^4 x + cos^4x = 1/16 + 9/16 = 10/16 = 5/8x=2π3−→sin4x+cos4x=116+916=1016=58
