1(sin4x+cos4x)2-2sin4xcos4x=17/32
{1-2sin2xcos2x}2-2sin4xcos4x=17/32
let sin2xcos2x=y
now,
{1-2y}2-2y2=17/32
2y2+1-4y=17/32
2y2-4y+15/32=0
now....y=1/8
hence sin2xcos2x=1/8
sin2x=1/√2
or,2x=45°
or.....x=22.5°
\hspace{-16}$Solve $\mathbf{\sin^8 x+\cos^8 x=\frac{17}{32}}$
-
UP 0 DOWN 0 0 1
1 Answers
1