Find a point for your equation by finding $f(x)$ at $\frac{\pi }{3}$

To find the slope, find $f'(x)$ and evaluate at $\frac{\pi }{3}$.

Find the equation of the tangent line by using the point-slope formula, $y-y_1 = m(x-x_1)$, and substituting $(x_1, y_1)$ and '$m$' with the values and slope you found in Hint 1 and Hint 2.

$f'(x)=\text{cos}x,$ $\ f\left ( \frac{\pi }{3} \right )=\text{sin}\frac{\sqrt{3}}{2},$ $\ f'\left ( \frac{\pi }{3} \right )=\text{cos}\frac{\pi }{3}= \frac{1}{2},$ tangent line is $y-\frac{\sqrt{3}}{2}=\frac{1}{2}\left ( x-\frac{\pi }{3} \right )$ or $y= \frac{1}{2}x+\frac{\sqrt{3}}{2}-\frac{\pi }{6}$.