Hello! Can someone please check and see if I did this right? Thanks! :)

Directions: Find the exact solutions of the equation in the interval [0,2pi]

cos2x+sinx=0

My answer:

cos2x+sinx=cos^2x-sin^2x+sinx
=1-sin^2x-sin^2x+sinx
=-2sin^2x+sinx+1=0

that is not an answer. It is a restatement of the problem, using only sines.

2sin^2x-sinx-1 == 0
(2sinx+1)(sinx-1) = 0
sinx = -1/2 or 1, so
x = π/2, 4π/3, 5π/3

http://www.wolframalpha.com/input/?i=cos2x%2Bsinx%3D0+for+x%3D0..2pi

To find the exact solutions of the equation cos2x + sinx = 0 in the interval [0, 2pi], you made a good start by using the trigonometric identity cos^2x = 1 - sin^2x to rewrite the equation. However, the next steps need to be corrected.

Let's go through the process again:

1. Start with the equation cos2x + sinx = 0.
2. Use the double angle formula for cos2x to rewrite cos2x as 1 - 2sin^2x.
The equation becomes 1 - 2sin^2x + sinx = 0.
3. Rearrange the equation to get the quadratic form: -2sin^2x + sinx + 1 = 0.

Now, we can solve this quadratic equation for sinx using factoring or the quadratic formula:

Factoring method:
1. Rearrange the equation to bring all terms on one side: -2sin^2x + sinx + 1 = 0.
2. Try to factorize the equation: (2sinx + 1)(-sinx + 1) = 0.
3. Set each factor equal to 0 and solve for sinx:
2sinx + 1 = 0 --> sinx = -1/2,
-sinx + 1 = 0 --> sinx = 1.

Using the Quadratic Formula:
1. Rearrange the equation to bring all terms on one side: -2sin^2x + sinx + 1 = 0.
2. Apply the quadratic formula: sinx = (-b ± √(b^2 - 4ac)) / (2a).
Here, a = -2, b = 1, and c = 1.
sinx = (-1 ± √(1 - 4(-2)(1))) / (2(-2)).
Simplify the equation: sinx = (-1 ± √9) / (-4).
3. Break down the equation into two solutions:
sinx = (-1 + 3) / (-4) --> sinx = 1.
sinx = (-1 - 3) / (-4) --> sinx = -1/2.

Now, it's time to check if these solutions fall within the given interval [0, 2pi]. You can substitute each value of sinx back into the original equation cos2x + sinx = 0 and see if it satisfies it.

Hope this clears things up!