Find all solutions in the interval [0,2pi): sin (cos x) = 0

*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.

## Comments

^{-1 ( x ) =0}