Q13/074) solve 2cos(x)≤∣ √∣1+sin(2x) ∣-√∣1-sin(2x) ∣∣≤√2

one solution cos x <= 0

so x is between pi/2 and - 3pi/2

second case cos x > = 0 and sin x >= 0

now sin x < cos x gives
cos x < 1/2 | cos x + sin x - ( cos x - sin x) <= 1/ sqrt(2)

or cos x < sin x <= 1/sqrt(2)

cos x >= 1/sqrt(2) => sin x >= 1/ sqrt(2) so no solution

cos x > 0 and sin x >= cos x gives cos x <= sqrt(2)

siimiliarly ranges sin x < 0 2 ranges | sin x | < cos x and | sin x | > cos x need to be anlaysed

by symetry we get cos <= 1/ sqrt(2)

so solution set cos^1 (1/ 2sqrt(2) to 2pi - cos^1 (1/2sqrt(2))