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

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

### 2 Answers by Expert Tutors

Philip P. | Effective and Affordable Math TutorEffective and Affordable Math Tutor
5.0 5.0 (438 lesson ratings) (438)
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

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.
Parviz F. | Mathematics professor at Community CollegesMathematics professor at Community Colle...
4.8 4.8 (4 lesson ratings) (4)
-1
Sin ( Cos X ) =0

√( 1 -Sin2 X ) =0

1 - Sin 2X = 0

Sin2 X = 1

SinX = ±1

X = ∏/2, 3∏/2

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