What value(s) of theta solve the following equation? cos^2(theta)-cos(theta)-6=0?
I try plugging in different numbers but I am not sure exactly what I am looking for in order to solve?
cos^2 Ø - cosØ - 6 = 0
let cosØ = x , then we get
x^2 - x - 6 = 0
(x-3)(x+2) = 0
x = 3 or x = -2
then cosØ = 3 or cosØ = -2
both of these are not possible, since the cosine is defined only between -1 and +1
there is no solution to your equation
also, you coul dhave reasoned thus:
cos^2θ <= 1
-1 <= cosθ <= 1
no way could you have cos^2θ - cosθ = 6
To solve the equation cos^2(theta) - cos(theta) - 6 = 0, you can use a quadratic equation.
1. Let's rewrite the equation as a quadratic equation by making a substitution. Let u = cos(theta), so the equation becomes u^2 - u - 6 = 0.
2. Now it's a quadratic equation in terms of u. To solve the quadratic equation, you can either factor it or use the quadratic formula.
Factoring:
u^2 - u - 6 = 0
(u - 3)(u + 2) = 0
Setting each factor equal to zero:
u - 3 = 0 or u + 2 = 0
Solving for u:
u = 3 or u = -2
3. Recall that u = cos(theta). So the values of u that solve the equation are 3 and -2.
4. Now we need to find the values of theta that correspond to the values of u. To find theta, you need to use the inverse cosine function (cos^(-1)).
theta = cos^(-1)(3)
theta = cos^(-1)(-2)
5. But before finding the values of theta, it's important to note that the domain of the inverse cosine function is -1 to 1. This means that there are no real values of theta that make cos(theta) equal to 3 or -2.
6. Therefore, the equation cos^2(theta) - cos(theta) - 6 = 0 has no real solutions for theta.