1. **State the problem:** Find the equation of the tangent line to the curve $f(x) = 5e^x + 2\sin x$ at $x=0$. 2. **Recall the formula for the tangent line:** The equation of the tangent line to $f(x)$ at $x=a$ is given by: $$y = f(a) + f'(a)(x - a)$$ where $f'(a)$ is the derivative of $f(x)$ evaluated at $x=a$. 3. **Find $f(0)$:** $$f(0) = 5e^0 + 2\sin 0 = 5 \times 1 + 2 \times 0 = 5$$ 4. **Find the derivative $f'(x)$:** $$f'(x) = \frac{d}{dx}(5e^x) + \frac{d}{dx}(2\sin x) = 5e^x + 2\cos x$$ 5. **Evaluate $f'(0)$:** $$f'(0) = 5e^0 + 2\cos 0 = 5 \times 1 + 2 \times 1 = 7$$ 6. **Write the tangent line equation at $x=0$:** $$y = f(0) + f'(0)(x - 0) = 5 + 7x$$ **Final answer:** The equation of the tangent line is $$y = 5 + 7x$$.