Find all values of x in the interval [0, 2π] that satisfy the equation. (Enter your answers as a comma-separated list.)
18 + 9 cos(2x) = 27 cos(x)
9cos2x - 27cosx + 18 = 0
9(2cos^2x-1)-27cosx+18 = 0
18cos^2x - 27cosx + 9 = 0
2cos^2x - 3cosx + 1 = 0
(2cosx-1)(cosx-1) = 0
cosx = 1/2 or 1
Take it from there.
To find all values of x that satisfy the equation, let's start by simplifying it.
18 + 9 cos(2x) = 27 cos(x)
First, distribute the 9 to cos(2x):
18 + 9 cos(2x) = 27 cos(x)
18 + 9(2cos^2(x) - 1) = 27 cos(x)
Next, distribute the 9 to both terms inside the parentheses:
18 + 18 cos^2(x) - 9 = 27 cos(x)
Combine like terms:
18 cos^2(x) - 9 = 27 cos(x) - 18
Now, rearrange the equation to form a quadratic equation:
18 cos^2(x) - 27 cos(x) + 9 = 0
To solve this quadratic equation, let's factor it:
(2 cos(x) - 1)(9 cos(x) - 9) = 0
Now, set each factor equal to zero and solve for x:
2 cos(x) - 1 = 0 or 9 cos(x) - 9 = 0
For the first equation:
2 cos(x) - 1 = 0
2 cos(x) = 1
cos(x) = 1/2
The solutions for x in the interval [0, 2π] are π/3 and 5π/3.
For the second equation:
9 cos(x) - 9 = 0
9 cos(x) = 9
cos(x) = 1
The solution for x in the interval [0, 2π] is 0.
Therefore, the values of x in the interval [0, 2π] that satisfy the equation are π/3, 5π/3, and 0.
The answer is π/3, 5π/3, 0.
To find all values of x in the interval [0, 2π] that satisfy the equation 18 + 9 cos(2x) = 27 cos(x), we can follow these steps:
1. Start by rearranging the equation to gather terms and get all the trigonometric functions on one side:
9 cos(2x) - 27 cos(x) + 18 = 0.
2. Next, we can use a trigonometric identity to rewrite cos(2x) in terms of x:
cos(2x) = 2 cos^2(x) - 1.
3. Substituting the value of cos(2x) in the equation, we get:
9(2 cos^2(x) - 1) - 27 cos(x) + 18 = 0.
4. Simplifying further, we have:
18 cos^2(x) - 9 - 27 cos(x) + 18 = 0.
5. Combining like terms gives:
18 cos^2(x) - 27 cos(x) + 9 = 0.
6. Now, we can try to factorize the quadratic equation. Recognizing that 9 = 3^2 and the leading coefficient is twice the constant term, we can factorize it as follows:
(3 cos(x) - 3)(6 cos(x) - 3) = 0.
7. Setting each factor equal to zero, we get two separate equations:
3 cos(x) - 3 = 0 --> cos(x) = 1,
and
6 cos(x) - 3 = 0 --> cos(x) = 1/2.
8. Solving the first equation cos(x) = 1, we find that x = 0 and x = 2π are solutions within the interval [0, 2π].
9. For the second equation cos(x) = 1/2, we need to find the angles in the interval [0, 2π] where cos(x) equals 1/2. These angles are π/3 and 5π/3.
10. Therefore, the values of x in the interval [0, 2π] that satisfy the equation 18 + 9 cos(2x) = 27 cos(x) are x = 0, x = 2π, x = π/3, and x = 5π/3.
Thus, the solution is x = 0, 2π, π/3, 5π/3.