Finding the solution sin and pi? Three questions

Question

1.Find the solution to the equation,

  sin x - 1 = 0 on 0 ≤ x < 2pi

2.Find all solutions of the equation 2 cos x - 1 = 0 on 0 ≤ x < 2pi

The answers are A and B, where A < B

A=

B=

3.Solve cos(x) = -0.54 on 0 ≤ x < 2pi

There are two solutions, A and B, with A < B

A=

B=

Answer 1

1.Given sin x - 1 = 0, we need to find what value x must be in order for, sin x = 1.

Notice that sin π ⁄ 2 = 1, therefore sin π ⁄ 2 - 1 = 0.

2.By rearranging the terms, we get cos x = 1 ⁄ 2.

This means that we need to find what values of x make cos x = 1 ⁄ 2.

From the unit circle, we can see that cos x = 1 ⁄ 2 when x = π ⁄ 3, or x = 5π ⁄ 3.

Therefore A = π ⁄ 3 and B = 5π ⁄ 3

3.We can solve for x by finding cos^-(0.54) = x.

For the first solution, A ≈ 122.68 degrees, and B can be obtained by B = 2π - A.

So B ≈ 237.32 degrees.