To solve the equation cos(3x) = √(-3/2), we can first find the cosine of 3x by using the identity cos(3x) = 4cos^3(x) - 3cos(x).
So, we have:
4cos^3(x) - 3cos(x) = √(-3/2).
Let's denote cos(x) as a, where -1 ≤ a ≤ 1. Substituting a into the equation, we get:
4a^3 - 3a = √(-3/2).
Now, we can square both sides to eliminate the square root:
(4a^3 - 3a)^2 = -3/2.
Expanding and simplifying, we get:
16a^6 - 24a^4 + 9a^2 = -3/2.
Rearranging terms, we have a 6th degree polynomial:
16a^6 - 24a^4 + 9a^2 + 3/2 = 0.
We can now solve this equation to find the possible values of a (cos(x)), which will lead us to the solutions for x.
To solve the equation cos(3x) = √(-3/2), we can first find the cosine of 3x by using the identity cos(3x) = 4cos^3(x) - 3cos(x).
So, we have:
4cos^3(x) - 3cos(x) = √(-3/2).
Let's denote cos(x) as a, where -1 ≤ a ≤ 1. Substituting a into the equation, we get:
4a^3 - 3a = √(-3/2).
Now, we can square both sides to eliminate the square root:
(4a^3 - 3a)^2 = -3/2.
Expanding and simplifying, we get:
16a^6 - 24a^4 + 9a^2 = -3/2.
Rearranging terms, we have a 6th degree polynomial:
16a^6 - 24a^4 + 9a^2 + 3/2 = 0.
We can now solve this equation to find the possible values of a (cos(x)), which will lead us to the solutions for x.