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

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2 Answers

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
 
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.
Sin ( Cos X ) =0
 
√( 1 -Sin2 X ) =0
 
 1 - Sin 2X = 0
 
Sin2 X = 1
 
SinX = ±1
 
  X = ∏/2, 3∏/2

Comments

Hi Parviz,
 
My calculator says sin(cos(57.2957..)) = 0.66958...
 
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 = n*pi but:
 
cos x = 0 at x = pi/2 and 3pi/2
 
Calculator confirms.
But for that should be   Sin ( Cos-1 ( x ) =0
 
       Sin ( cosX ) =0
 
     -1<cosx<+1
   
   CosX = 0  is the only acceptible answer
 
     in which case X = ∏/2, 3∏/2.
   Thank you. I made the correction. My mistake was I took the small angle where SinX/ X =1 x →0      

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