Solve the following equation for all solutions: cos(5x)cos(3x)-sin(5x)sin(3x)=√3/2. I know the formula to use would be cos(A+B), but I'm just not confident in my ability to solve it correctly.
cos(5x)cos(3x)-sin(5x)sin(3x)=√3/2
cos(5x+3x) = √3/2
8x = ±π/6 + 2kπ
x = ±π/48 + kπ/4
To solve the equation cos(5x)cos(3x) - sin(5x)sin(3x) = √3/2, we can apply the angle-sum identity for cosine:
cos(A + B) = cos(A)cos(B) - sin(A)sin(B)
In this case, let's consider A = 5x and B = 3x. So the equation becomes:
cos(5x + 3x) = √3/2
Simplifying, we have:
cos(8x) = √3/2
Now, we need to find the values of 8x that satisfy this equation. Using the inverse cosine function (arccos), we can write:
8x = arccos(√3/2)
To find the values of x, we need to consider the range of the inverse cosine function, which is between 0 and π. Since we have 8x on the left side, we can multiple both sides by 1/8:
x = (1/8) * arccos(√3/2)
So, the solutions for x are given by:
x = (1/8) * arccos(√3/2) + (k * π/8)
Where k is an integer.
To solve the given equation, we can use the trigonometric identity:
cos(A + B) = cos(A)cos(B) - sin(A)sin(B)
Let's convert the given equation using this identity:
cos(5x)cos(3x) - sin(5x)sin(3x) = √3/2
cos(5x + 3x) = √3/2
cos(8x) = √3/2
Now, to find the solutions, we need to know the values of x for which cos(8x) is equal to √3/2.
First, we can find the principal value of 8x using the inverse cosine function:
8x = arccos(√3/2)
Next, we need to find the general solution for 8x. To do that, we can use the periodicity of the cosine function.
Since the cosine function has a period of 2π, we can add multiples of 2π to the principal value to find other solutions.
The general solution for 8x is:
8x = arccos(√3/2) + 2kπ
where k is an integer.
Finally, you can solve for x by dividing both sides of the equation by 8:
x = (arccos(√3/2) + 2kπ) / 8
This gives you the general solution for x. To find all solutions within a specific interval, you can substitute different integer values of k and solve for x.