F.4 trigonometry questions

1. sin (90 + x) = cos x
tan (90 + x) = -1/tan x
cos (180 + x) = - cos x
cos (-x) = cos x
sin (270 + x) = - cos x
so LHS = ( - cos x tan x - cos x)^2 + (cos x - cos x tan x)^2 = ( - sin x - cos x)^2 + (cos x - sin x)^2 = (sin x + cos x)^2 + (cos x - sin x)^2
= sin^2 x + 2 sin x cos x + cos^2 x + cos^2 x - 2 sin x cos x + sin^2 x
= 2 sin^2 x + 2 cos^2 x = 2(sin^2 x + cos^2 x) = 2 = RHS.
2.
cos x (sin x - k) = 0
cos x = 0 ..............(1)
sin x - k = 0 ..............(2)
For 0 < x < 360
From (1)
x = 90, 270.
From (2)
x = arc sin (k)
Since k is a positive real number, x will have 2 distinct roots, one in the 1st Quadrant and the other in the 2nd Quadrant.
Together with 90 or 270, the equation will have 3 roots. Do not understand why there could be 2 distinct roots only. Please clarify the meaning of the question.
(if really 2 distinct roots, it can only be 90 and 270, so k = 1 only.)

If cos x sin (x-k) = 0 has only two distinct solutions, then k can only be 90° ?