Philip P. answered 04/13/14
Tutor
4.9
(484)
Effective and Patient Math Tutor
Find all solutions in the interval [0,2pi): sin (cos x) = 0
sin x = 0 when x = 0, pi, and 2pi on the interval [0,2pi]
So sin(cos x) = 0 when cos x = 0, pi, 2pi, ...
The range of cos x is ±1, so it can't = pi or 2pi, but:
cos x = 0 at x = pi/2 and 3pi/2
So sin(cos x) = 0 when cos x = 0, pi, 2pi, ...
The range of cos x is ±1, so it can't = pi or 2pi, but:
cos x = 0 at x = pi/2 and 3pi/2
Hence sin(cos(pi/2)) = sin(cos(3pi/2) = 0
Hence the solutions on the interval [0,2pi] are pi/2 and 3pi/2 radians
Calculator confirms.
Calculator confirms.
Philip P.
04/12/14