Solve the given equation. (Enter your answers as a comma-separated list. Let k be any integer. Round terms to three decimal places where appropriate. If there is no solution, enter NO SOLUTION.)
2 cos2 θ − cos θ − 1 = 0
To solve the given equation, we can use a substitution to simplify it. Let's substitute u = cos(θ):
2u^2 - u - 1 = 0
Now we have a quadratic equation in terms of u. We can solve this equation by factoring, completing the square, or using the quadratic formula.
In this case, let's use the quadratic formula: u = (-b ± √(b^2 - 4ac)) / 2a
For the given equation, a = 2, b = -1, and c = -1. Plugging these values into the quadratic formula:
u = (1 ± √((-1)^2 - 4 * 2 * (-1))) / (2 * 2)
u = (1 ± √(1 + 8)) / 4
u = (1 ± √9) / 4
u = (1 ± 3) / 4
This gives us two possible values for u:
u1 = (1 + 3) / 4 = 4 / 4 = 1
u2 = (1 - 3) / 4 = -2 / 4 = -0.5
Now we can substitute back to find the values of θ:
For u1 = 1, cos(θ) = 1
θ = arccos(1) = 0
For u2 = -0.5, cos(θ) = -0.5
θ = arccos(-0.5) = π ± arccos(0.5)
Therefore, the solutions for the equation are θ = 0 and θ = π ± arccos(0.5).
Note: It is important to keep in mind that the range of the arccosine function is restricted to values between 0 and π radian (or 0 and 180 degrees), so make sure to consider the correct interval for θ based on the context of the problem.
what's the problem. You know how to solve quadratics:
Let x = cosθ
2x^2 - x - 1 = 0
(2x+1)(x-1) = 0
x = -1/2 or 1
So,
cosθ = -1/2 or 1
. . .