Answer:
To find the equation of the tangent line to the graph of the function \(f(x) = 2\cos(x) - 3\sin(x)\) at the point \((\pi, -2)\), you'll need to find both the slope of the tangent line and the y-intercept.
The slope of the tangent line at a point can be found by taking the derivative of the function and then evaluating it at the given x-coordinate.
First, let's find the derivative of \(f(x)\):
\[f'(x) = \frac{d}{dx}(2\cos(x) - 3\sin(x))\]
Using the derivative rules:
\[f'(x) = -2\sin(x) - 3\cos(x)\]
Now, we need to find the slope of the tangent line at \(x = \pi\):
\[f'(\pi) = -2\sin(\pi) - 3\cos(\pi) = 0 - (-3) = 3\]
So, the slope of the tangent line at \((\pi, -2)\) is 3.
Now that we have the slope (\(m = 3\)), we can use the point-slope form of a linear equation to find the equation of the tangent line:
\[y - y_1 = m(x - x_1)\]
Where \((x_1, y_1)\) is the given point on the curve \((\pi, -2)\). Plugging in the values:
\[y - (-2) = 3(x - \pi)\]
Now, simplify the equation:
\[y + 2 = 3x - 3\pi\]
To express it in slope-intercept form (\(y = mx + b\)), isolate \(y\):
\[y = 3x - 3\pi - 2\]
So, the equation of the tangent line to the graph of \(f(x) = 2\cos(x) - 3\sin(x)\) at the point \((\pi, -2)\) is:
\[y = 3x - 3\pi - 2\]
Explanation: