How many solutions does the equation sin(`pi/2cosx) = cos( pi /2sinx) ` have in the close interval [0,`pi] ?`

The solution of the equation `cos(pi/2*sin x) = sin(pi/2*cos x)` in `[0, pi]` is `(0, pi/2)`.

Expert Answers

An illustration of the letter 'A' in a speech bubbles

The solutions of `cos(pi/2*sin x) = sin(pi/2*cos x)` have to be determined.

`cos(pi/2*sin x) = sin(pi/2*cos x)`

`=> sin (pi/2 - pi/2*sin x) = sin(pi/2*cos x)`

`=> pi/2 - pi/2*sin x = pi/2*cos x` `pi/2 - pi/2*sin x = -pi/2*cos x`

`pi/2 - pi/2*sin x = pi/2*cos x`

=> `1 - sin x = cos x`

The solution of `1 - sin x - cos x = 0` can be determined by looking at the graph of the function.

The solution of `1- sin x - cos x = 0` in `[0,pi]` is `0` and `pi/2`.

The solution of the equation `cos(pi/2*sin x) = sin(pi/2*cos x)` in `[0, pi]` is `(0, pi/2)`.

Last Updated by eNotes Editorial on February 23, 2021
Soaring plane image

We’ll help your grades soar

Start your 48-hour free trial and unlock all the summaries, Q&A, and analyses you need to get better grades now.

  • 30,000+ book summaries
  • 20% study tools discount
  • Ad-free content
  • PDF downloads
  • 300,000+ answers
  • 5-star customer support
Start your 48-Hour Free Trial