Use a trig identity to combine two functions into one so you can solve for x. (The solution should be valid for any value of t).

3cos(t) + 3*sqrt(3)*sin(t)=6cos(t-x)

I know that 6 cos(t-x) can be 6(cos(t)cosx(x)+sin(t)sin(x))
I don't know where to go from there though.

what you have now is

3cost + 3√3 sint = 6cosx cost + 6sinx sint

for them to be identical, the coefficients must match:

3 = 6cosx
3√3 = 6sinx

looks like x = π/3

To combine the functions on the left side of the equation into one trigonometric function, we can use the identity:

cos(a - b) = cos(a)cos(b) + sin(a)sin(b)

In your equation, you have 6cos(t - x) on the right side, which can be written as 6(cos(t)cos(x) + sin(t)sin(x)).

Now, we can equate the two expressions:

3cos(t) + 3sqrt(3)sin(t) = 6(cos(t)cos(x) + sin(t)sin(x))

To solve for x, we need to eliminate the t terms. This can be achieved by comparing the coefficients of cos(t) and sin(t) on both sides of the equation.

Comparing the coefficients:

3 = 6cos(x)
3sqrt(3) = 6sin(x)

Dividing both sides of the first equation by 6:

1/2 = cos(x)

Dividing both sides of the second equation by 6 and sqrt(3):

1/2 = sin(x)/sqrt(3)

Now, we can use the Pythagorean identity to relate sin(x) and cos(x):

sin^2(x) + cos^2(x) = 1

Substituting the values we have obtained:

(1/2)^2 + (cos(x))^2 = 1
1/4 + (cos(x))^2 = 1
(cos(x))^2 = 1 - 1/4
(cos(x))^2 = 3/4

Taking the square root of both sides:

cos(x) = sqrt(3)/2

So, x can be any angle where the cosine is equal to sqrt(3)/2. One such angle is x = π/6 (30 degrees).

Therefore, the solution for x is x = π/6 + 2kπ, where k is an integer. This solution is valid for any value of t.