Find an equation of the tangent line to $y = \cos(x)+3\sin(x)$ at $x=\pi/3$ 
Find an equation of the tangent line to 
  $$y = \cos(x)+3\sin(x)$$ at $x=\pi/3$.

This is what I have done...
Find $y$, $y= \cos(\pi/3) + 3\sin(\pi/3)$
this equals $1 + \sqrt 3/2$
Next
Find $f'(x) = \sin(\pi/3) + 3\cos(\pi/3)$
this equals $3+ 3\sqrt3/2$
Next
Plug into point slope form
$(y-1+\sqrt 3/2) = (3 +3\sqrt3/2) (x - \pi/3)$
$y = \left(\frac{3+3\sqrt 3}{2}\right)(x-\pi/3)+(1+3\sqrt 3/2)$
Am I doing something wrong? Thanks. 
 A: While your strategy is correct, you incorrectly evaluated the sine and cosine functions at $x = \pi/3$ and incorrectly took the derivative.
Let $f(x) = \cos x + 3\sin x$.  Since 
$$\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$$
and 
$$\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$$
when we evaluate $f(x)$ at $\pi/3$ we obtain

 $$f\left(\frac{\pi}{3}\right) = \cos\left(\frac{\pi}{3}\right) + 3\sin\left(\frac{\pi}{3}\right) = \frac{1}{2} + \frac{3\sqrt{3}}{2} = \frac{1 + 3\sqrt{3}}{2}$$

Since the derivative of $g(x) = \cos x$ is $g'(x) = -\sin x$ and the derivative of $h(x) = \sin x$ is $h'(x) = \cos x$, the derivative of $f(x)$ is 

 $$f'(x) = -\sin x + 3\cos x$$

Thus, the derivative of $f(x)$ evaluated at $\pi/3$ is 

 $$f'\left(\frac{\pi}{3}\right) = -\sin\left(\frac{\pi}{3}\right) + 3\cos\left(\frac{\pi}{3}\right) = -\frac{\sqrt{3}}{2} + \frac{3}{2} = \frac{3 - \sqrt{3}}{2}$$

Hence, the equation of the tangent line is 

 $$y - \frac{1 + 3\sqrt{3}}{2} = \frac{3 - \sqrt{3}}{2}\left(x - \frac{\pi}{3}\right)$$

