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.