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

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

Calculator confirms.

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

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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.

Parviz F. | Mathematics professor at Community CollegesMathematics professor at Community Colle...

Sin ( Cos X ) =0

√( 1 -Sin^{2 }X ) =0

1 - Sin ^{2}X = 0

Sin^{2} X = 1

SinX = ±1

X = ∏/2, 3∏/2

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## Comments

^{-1 ( x ) =0}