What values of theta between 0 and 2pi solve the following equation?

sin squared (2theta)-sin(2theta)-2=0

NOTE: if you choose to plot this, a good WINDOW is (0,2pi) x (-3,3).

if you let x=sin(2θ) you have

x^2-x-2=0
(x-2)(x+1) = 0
so,
sin(2θ) = -1
2θ = 3π/2,7π/2
θ = 3π/4,7π/4

To solve the equation sin^2(2θ) - sin(2θ) - 2 = 0, we can use a substitution method. Let's substitute sin(2θ) with x, so the equation becomes:

x^2 - x - 2 = 0

Now we can solve this quadratic equation. We can either factor it or use the quadratic formula.

Factoring:

(x + 1)(x - 2) = 0

Setting each factor equal to zero:

x + 1 = 0 or x - 2 = 0

Solving for x:

x = -1 or x = 2

Since we substituted sin(2θ) with x, we can now set these two values for x equal to sin(2θ) and solve for θ.

For x = -1, we have sin(2θ) = -1. Solving for θ, we find:

2θ = arcsin(-1) = -π/2 + 2πk, where k is an integer.

Dividing by 2:

θ = (-π/2 + 2πk) / 2

Simplifying:

θ = -π/4 + πk, where k is an integer.

For x = 2, we have sin(2θ) = 2, which is impossible since the range of sine function is -1 to 1. Therefore, x = 2 is not a valid solution for sin(2θ).

Therefore, the values of θ between 0 and 2π that solve the equation sin^2(2θ) - sin(2θ) - 2 = 0 are:

θ = -π/4 + πk, where k is an integer.

If you would like to plot this equation on a graphing calculator or software, a good window to use is (0, 2π) for the x-axis and (-3, 3) for the y-axis.