Solving 4x for sin^-1(1) between -pi and pi

[spynjr]

Homework Statement

Find all angles x, -pi <= x <= pi which satisfy sin4x = 1

Homework Equations

The Attempt at a Solution

Solved x, ie

4x = sin^-1(1)
4x = pi/2
x = pi/8

Sine is positive for angles in quadrants one and two, so there's one other angle which is in quadrant two?

x = pi - pi/8
x = 7pi/8

Final answer, x = pi/8, 7pi/8

Just not feeling confident. There are about five other similar questions, ie solving x between -pi and pi

 

[carrotcake10]

Well, you could always just plug your solutions into a calculator. :)

But really there is only 1 point on the circle that sin(x) = 1. This is at pi/2. But the function you are given multiplies whatever argument you pass to sin by 4. In order for this to come out correctly, you need to divide pi/2 by 4.

Your work looks ok, but there isn't a solution in quadrant 2.

 

[Mark44]

Homework Statement

Find all angles x, -pi <= x <= pi which satisfy sin4x = 1

Homework Equations

The Attempt at a Solution

Solved x, ie

4x = sin^-1(1)
4x = pi/2
x = pi/8

Sine is positive for angles in quadrants one and two, so there's one other angle which is in quadrant two?

x = pi - pi/8
x = 7pi/8

Final answer, x = pi/8, 7pi/8

Just not feeling confident. There are about five other similar questions, ie solving x between -pi and pi

x = pi/8 is a solution, but 7pi/8 is not.

Since your solutions have to satisfy -pi <= x <= pi, then -4pi <= 4x <= 4pi. There are two values of 4x for which sin(4x) = 1, of which one of them is pi/2, hence x = pi/8, which you've already gotten. What is the other value of 4x in [0, 4pi] for which sin(4x) = 1? Once you get it, you can solve for x.

Also, there are two values of 4x in [-4pi, 0] for which sin(4x) = 1. What are they? Once you get them, solve for x for each of them.