Calculus BC problem So I as attempting to solve this problem and just got stuck overall. Since this is just practice an explanation is much more valuable than the answer itself. Thanks!
Here is the problem

 A: Hint:
The slope of the line is $m=\dfrac{\sin(K)-0}{K-0}=\dfrac{\sin(K)}K$, then its equation is $$y=\dfrac{\sin(K)}Kx$$
Now, observe that
\begin{align*}\text{Yellow area }&=\int_0^K\left(\sin x-\dfrac{\sin(K)}Kx\right)dx\\&=1-\cos(K)-\frac{\sin(K)}{2K}K^2\\&=1-\cos(K)-\tfrac K2\sin(K)
\end{align*}
And
\begin{align*}
\text{Green area }&=\int_0^K\dfrac{\sin(K)}Kx\,dx+\int_K^{\pi}\sin x\,dx\\
&=\tfrac K2\sin(K)-\cos\pi+\cos(K)\\
&=1+\cos(K)+\tfrac K2\sin(K)
\end{align*}
A: $$\int_0^\pi \sin(x)dx =- cos(x)[_0^\pi =2 $$
So the green part must have an area of 1 
Split the green part into ...
a right triangle with base $b=k$ and height $ h=sin(k)$ 
and
the area under the curve between $x=k$ and $x = \pi $
$$ \frac 12 k \sin(k) + \int_k^\pi \sin(x) dx = 1  $$
$$ \frac 12 k \sin(k) + 1-\cos(k) = 1  $$
A: We have $\displaystyle \int_0^{π}\sin(x)\,dx=2$.
The slope of the line is $\displaystyle \frac{\sin(K)}{K}.$
Therefore, $\displaystyle \int_0^{K}\left(\sin(x)-\frac{x\sin(K)}{K}\right)=1$ (Yellow Area)
This means $\displaystyle \left(-\cos(k)-\frac{1}{2}\frac{x^2\sin(K)}{K}\right)\biggr\rvert_0^K=-\cos(K)+1-\frac{1}{2}K\sin(K)=1$.
$\displaystyle \cos(K)+\frac{1}{2}K\sin(K)=0$.
$\boxed{K\sin(K)+2\cos(K)=0}$.

Answer choice E.
