Finding values of $\theta$ for which $(\cos\theta,\sin\theta)$ lies inside the triangle formed by $x+y=2$, $x−y=1$, and $6x+2y−\sqrt{10}​=0$ 
Find the range of values of $\theta$, such that $\theta\in[0,2\pi]$ for which $(\cos \theta,\sin \theta)$ lies inside the triangle formed by $x + y = 2, x − y = 1$ and $6x + 2y − \sqrt{10} ​=0$.

Not getting any hint of how can I get the range of those points is there any short method of doing so.
I have plotted the graph and then I found out that the line $6x + 2y − \sqrt{10} ​=0$ which forms one of the sides of the triangle has $O(0,0)$ and $P(\cos \theta, \sin\theta)$ on opposite sides so applied the formula of power of point.
So $L_3(O)\cdot L_3(P)<0$ and then solved to find out the range.
If there is any short method please tell me.


Working from the diagram OP has posted, $D=(1,0)$ corresponds to $\theta=0$, so we only need to work out $E$. Now, $E$ is the intersection of $6x+2y=\sqrt{10}$ and $x^2+y^2=1$. In polar coordinates, these are $6r\cos\theta+2r\sin\theta=\sqrt{10}$ and $r=1$, so it comes down to solving $6\cos\theta+2\sin\theta=\sqrt{10}$. Noting that $6^2+2^2=40$, we divide both sides by $\sqrt{40}=2\sqrt{10}$ to get $${3\over\sqrt{10}}\cos\theta+{1\over\sqrt{10}}\sin\theta={1\over2}$$ Let $\eta=\arcsin{3\over\sqrt{10}}$ which implies $\cos\eta={1\over\sqrt{10}}$ and $\tan\eta=3$. Then we have $1/2=\sin\eta\cos\theta+\cos\eta\sin\theta=\sin(\theta+\eta)$. This is satisfied by $\theta+\eta=\pi/6$, but $\tan\eta=3>1$ implies $\eta>\pi/4$ and thus $\theta+\eta>\pi/4$, so we go to $\theta+\eta=5\pi/6$. Then, $$\theta={5\pi\over6}-\eta={5\pi\over6}-\arctan3$$
