$$\sum {P(X) = 1 \Rightarrow k + 2k + 3} k + k = 1$$

$$ \Rightarrow k = {1 \over 9}$$

Now, $$p = P\left( {{{kx < 4} \over {X < 3}}} \right) = {{P(X = 2)} \over {P(X < 3)}} = {{{{2k} \over {9k}}} \over {{k \over {9k}} + {{2k} \over {9k}}}} = {2 \over 3}$$

$$ \Rightarrow p = {2 \over 3}$$

Now, $$5p = \lambda k$$

$$ \Rightarrow (5)\left( {{2 \over 3}} \right) = \lambda (1/9)$$

$$ \Rightarrow \lambda = 30$$

Given equation

$${\sin ^4}\theta + {\cos ^4}\theta - \sin \theta \cos \theta = 0$$

$$ \Rightarrow 1 - {\sin ^2}\theta {\cos ^2}\theta - \sin \theta \cos \theta = 0$$

$$ \Rightarrow 2 - {(\sin 2\theta )^2} - \sin 2\theta = 0$$

$$ \Rightarrow {(\sin 2\theta )^2} + (\sin 2\theta ) - 2 = 0$$

$$ \Rightarrow (\sin 2\theta + 2)(\sin 2\theta - 1) = 0$$

$$ \Rightarrow \sin 2\theta = 1$$ or $$\sin 2\theta = - 2$$ (Not Possible)

$$ \Rightarrow 2\theta = {\pi \over 2},{{5\pi } \over 2},{{9\pi } \over 2},{{13\pi } \over 2}$$

$$ \Rightarrow \theta = {\pi \over 4},{{5\pi } \over 4},{{9\pi } \over 4},{{13\pi } \over 4}$$

$$ \Rightarrow S = {\pi \over 4} + {{5\pi } \over 4} + {{9\pi } \over 4} + {{13\pi } \over 4} = 7\pi $$

$$ \Rightarrow {{8S} \over \pi } = {{8 \times 7\pi } \over \pi } = 56.00$$

Case I : When cot x > 0, $$x \in \left[ {0,{\pi \over 2}} \right] \cup \left[ {\pi ,{{3\pi } \over 2}} \right]$$

$$\cot x = \cot x + {1 \over {\sin x}} \Rightarrow $$ not possible

Case II : When cot x < 0, $$x \in \left[ {{\pi \over 2},\pi } \right] \cup \left[ {{{3\pi } \over 2},2\pi } \right]$$

$$ - \cot x = \cot x + {1 \over {\sin x}}$$

$$ \Rightarrow {{ - 2\cos x} \over {\sin x}} = {1 \over {\sin x}}$$

$$ \Rightarrow \cos x = {{ - 1} \over 2}$$

$$ \Rightarrow x = {{2\pi } \over 3},{{4\pi } \over 3}$$(Rejected)

One solution.

We know,

$$ - \sqrt {{a^2} + {b^2}} \le a\cos x + b\sin x \le \sqrt {{a^2} + {b^2}} $$

$$ \therefore $$ $$ - \sqrt {{3^2} + {4^2}} \le 3\cos x + 4\sin x \le \sqrt {{3^2} + {4^2}} $$

$$ - 5 \le k + 1 \le 5$$

$$ - 6 \le k \le 4$$

$$ \therefore $$ Set of integers = $$ - 6, - 5, - 4, - 3, - 2, - 1,0,1,2,3,4$$ = Total 11 intergers.