Integral trigonometry

as, $-\sqrt{74}\leq 7\cos{x}+5\sin{x}\leq \sqrt{74}$ therefore $-\sqrt{74}\leq 2k+1\leq \sqrt{74}$ Hence $k=-4,-3,-2,-1,0,1,2,3$

Using Pythagoras theorem, we get:-

$\sqrt{5^{2}+7^{2}}= \sqrt{74}$ which is equal to $8.6$ and then minimum value should be $-8.6$

So the values of $k$ are ranged between $8.6$ and $-8.6$

Since $2k+1$ is an odd integer, therefore there are $8$ odd integers between $-8.6$ to $8.6$ i.e. $-7, -5, -3, -1, 1, 3, 5, 7$ ..

Find the value of $k$ by equating $2k+1$ to those odd integers and you'll notice that there lie $8$ values of $k$ between $-8.6$ and $8.6$ .