find the equation of the tangent line to the curve y = ln(3 - 2 cos(4x)) at (0,0).

Explanation:

Step1: Differentiate using chain - rule

Let $u = 3-2\cos(4x)$. Then $y=\ln(u)$. The derivative of $\ln(u)$ with respect to $u$ is $\frac{1}{u}$, and the derivative of $u = 3-2\cos(4x)$ with respect to $x$ is $8\sin(4x)$ (using the chain - rule on $\cos(4x)$). By the chain - rule $\frac{dy}{dx}=\frac{8\sin(4x)}{3 - 2\cos(4x)}$.

Step2: Find the slope at the given point

Substitute $x = 0$ into $\frac{dy}{dx}$. When $x = 0$, $\frac{dy}{dx}=\frac{8\sin(0)}{3-2\cos(0)}=\frac{0}{3 - 2\times1}=0$.

Step3: Use the point - slope form of a line

The point - slope form of a line is $y - y_1=m(x - x_1)$, where $(x_1,y_1)=(0,0)$ and $m = 0$. Substituting these values, we get $y-0=0(x - 0)$.

Answer:

$y = 0$