2cos^2Beta-Cosbeta=0 (find all solutions for the equation in the interval 0, 2 pi). I have no idea how to go about doing this. Do you isolate the beta on one side? Would you be able to show me how and I'll solve for you to check?

cos B (2 cos B - 1 )=0

cos B = 0 is one solution. That is at pi/2 and 3pi/2
cos B = 1/2 is also a solution
that is at 60 degrees and -60 degrees which is at pi/3 and 5 pi/3

I thought cos=1/2 at 30, not 60

To solve the equation 2cos^2(beta) - cos(beta) = 0 in the interval (0, 2pi), you can follow these steps:

Step 1: Simplify the equation
First, notice that the equation can be factored. Rewrite the equation as:
cos(beta)(2cos(beta) - 1) = 0

Now, we have two separate factors: cos(beta) = 0 and 2cos(beta) - 1 = 0.

Step 2: Solve for cos(beta) = 0
In the interval (0, 2pi), cos(beta) = 0 at two points: beta = pi/2 and beta = 3pi/2. These are solutions to the first factor.

Step 3: Solve for 2cos(beta) - 1 = 0
Now, solve the second factor, 2cos(beta) - 1 = 0.

Add 1 to both sides: 2cos(beta) = 1
Divide both sides by 2: cos(beta) = 1/2

In the interval (0, 2pi), cos(beta) = 1/2 at two points: beta = pi/3 and beta = 5pi/3. These are solutions to the second factor.

Step 4: Combine the solutions
To find all the solutions in the interval (0, 2pi), combine the solutions from Step 2 and Step 3:

beta = pi/2, 3pi/2, pi/3, 5pi/3

These are the four solutions to the given equation in the interval (0, 2pi).

Now, you can substitute these values of beta into the original equation to check if they satisfy the equation.

I hope this explanation helps you understand the process.