(6 pts) 7) Find the equation of the tangent line to the curve $\left\{\begin{array}{l}x=\sin 2 t+\cos t \\ y=\cos 2 t-\sin t\end{array}\right.$ at the point $t=\pi$.

Step 1 ：Given the parametric equations for x and y as functions of t, we have \(x = \sin(2t) + \cos(t)\) and \(y = \cos(2t) - \sin(t)\).

Step 2 ：We need to find the derivatives of these equations with respect to t. The derivative of x with respect to t is \(dx/dt = 2\cos(2t) - \sin(t)\) and the derivative of y with respect to t is \(dy/dt = -2\sin(2t) - \cos(t)\).

Step 3 ：The slope of the tangent line to the curve at a given point is equal to the derivative of y with respect to x, which is \(dy/dx = (dy/dt) / (dx/dt)\).

Step 4 ：Evaluating these derivatives at \(t = \pi\), we find that \(dx/dt = -1\) and \(dy/dt = -1\), so the slope of the tangent line is \(m = dy/dx = 1\).

Step 5 ：Substituting \(t = \pi\) into the original equations, we find that the coordinates of the point on the curve are \(x = -1\) and \(y = 1\).

Step 6 ：The equation of a line is given by \(y = mx + c\), where m is the slope and c is the y-intercept. Substituting the slope and the coordinates of the point into this equation, we can solve for c to find that \(c = y - mx = 1 - (-1) = 2\).

Step 7 ：\(\boxed{\text{Final Answer: The equation of the tangent line to the curve at the point } t=\pi \text{ is } y = x + 2}\)