Grove HSC 3U Ex 5.9 Question 4

Hello sir, how do you do question 4?

SOLUTION
Find $y'$, at $x=\pi/6$ and sub into $y-y_1=m(x-x_1)$

$y'=\cos(\pi-x)\times -1$
$y'=-\cos(\pi-x)$
So, tangent gradient is 
$m=y'(\pi/6)=-\cos (\pi-\pi/6)=-\cos(5\pi/6)=\cos(\pi/6)=\sqrt{3}/2$

Eqn of tangent

$y-0.5=\sqrt{3}/2(x-\pi/6)$
$2y-1=\sqrt{3}(x-\pi/6)$
$2y-1=\sqrt{3}x-\sqrt{3}\pi/6$
$12y-6=6\sqrt{3}x-\sqrt{3}\pi$
$6\sqrt{3}x-12y+6-\sqrt{3}\pi=0$

Thanks for the question.
Solution brought to you from the Jan Academy 2016, JSH :-)