Given f″(x)=−4sin(2x) and f′(0)=3 and f(0)=−6.

Find f(π/6)

f"(x) = -4sin(2x)

f'(x) = 2cos(2x) + c1
f(x) = sin(2x) + c1*x + c2

Now plug in the initial conditions to find c1 and c2

To find f(π/6), we will first integrate f″(x) and then apply the initial conditions f′(0) = 3 and f(0) = -6.

Step 1: Integrate f″(x) to find f′(x):

∫ f″(x) dx = -4∫ sin(2x) dx

Using the double-angle formula for sine:

= -2∫ sin(2x) dx = -2(-1/2)cos(2x) + C

= cos(2x) + C1

Here, C1 is the constant of integration.

Step 2: Apply the initial condition f′(0) = 3:

f′(0) = cos(2(0)) + C1 = 1 + C1 = 3

Solving for C1, we get:

C1 = 3 - 1 = 2

So, the antiderivative of f″(x) is f′(x) = cos(2x) + 2.

Step 3: Integrate f′(x) to find f(x):

∫ f′(x) dx = ∫ (cos(2x) + 2) dx = ∫ cos(2x) dx + ∫ 2 dx

= (1/2)sin(2x) + 2x + C2

Here, C2 is the constant of integration.

Step 4: Apply the initial condition f(0) = -6:

f(0) = (1/2)sin(2(0)) + 2(0) + C2 = 0 + 0 + C2 = C2 = -6

So, the antiderivative of f′(x) is f(x) = (1/2)sin(2x) + 2x - 6.

Step 5: Find f(π/6):

f(π/6) = (1/2)sin(2(π/6)) + 2(π/6) - 6

Simplifying further, we get:

= (1/2)sin(π/3) + (π/3) - 6

Using the trigonometric values of π/3:

= (1/2)(√3/2) + (π/3) - 6

= (√3/4) + (π/3) - 6

Hence, f(π/6) is (√3/4) + (π/3) - 6.

To find f(π/6), we need to integrate the given equation f″(x) = -4sin(2x) twice and use the initial conditions f′(0) = 3 and f(0) = -6.

First, integrate f″(x) = -4sin(2x) once to find f′(x):

∫ f″(x) dx = ∫ -4sin(2x) dx
f′(x) = -4 ∫ sin(2x) dx

Using the integral of sin(x), we have:

f′(x) = -4 * (-1/2) * cos(2x) + C
f′(x) = 2cos(2x) + C

Applying the initial condition f′(0) = 3, we can solve for the constant C:

f′(0) = 2cos(2(0)) + C
3 = 2cos(0) + C
3 = 2 * 1 + C
3 = 2 + C
C = 1

Therefore, we have:

f′(x) = 2cos(2x) + 1

Now, integrate f′(x) = 2cos(2x) + 1 once again to find f(x):

∫ f′(x) dx = ∫ (2cos(2x) + 1) dx
f(x) = 2 ∫ cos(2x) dx + ∫ dx

Using the integral of cos(x), we have:

f(x) = 2 * (1/2) * sin(2x) + x + D
f(x) = sin(2x) + x + D

Applying the initial condition f(0) = -6, we can solve for the constant D:

f(0) = sin(2(0)) + 0 + D
-6 = sin(0) + D
-6 = 0 + D
D = -6

Therefore, we have:

f(x) = sin(2x) + x - 6

To find f(π/6), substitute π/6 for x:

f(π/6) = sin(2(π/6)) + (π/6) - 6

Simplifying this expression will give us the value of f(π/6).