If $$\theta_1$$ and $$\theta_2$$ be respectively the smallest and the largest values of $$\theta$$ in $$(0, 2\pi) - \{\pi\}$$ which satisfy the equation, $$2\cot^2\theta - \frac{5}{\sin\theta} + 4 = 0$$, then $$\int_{\theta_1}^{\theta_2} \cos^2 3\theta \, d\theta$$ is equal to:

We have to solve the trigonometric equation $$2\cot^2\theta-\dfrac{5}{\sin\theta}+4=0$$ for $$\theta\in(0,2\pi)\setminus\{\pi\}$$, obtain the smallest root $$\theta_1$$ and the largest root $$\theta_2$$ in that interval, and then evaluate the definite integral $$\displaystyle\int_{\theta_1}^{\theta_2}\cos^2 3\theta\,d\theta$$.

First, recall that $$\cot\theta=\dfrac{\cos\theta}{\sin\theta}$$, so $$\cot^2\theta=\dfrac{\cos^2\theta}{\sin^2\theta}$$. Substituting this in the given equation we get

$$2\left(\dfrac{\cos^2\theta}{\sin^2\theta}\right)-\dfrac{5}{\sin\theta}+4=0.$$

To clear denominators we multiply every term by $$\sin^2\theta$$:

$$2\cos^2\theta-5\sin\theta+4\sin^2\theta=0.$$

Now we express $$\cos^2\theta$$ in terms of $$\sin\theta$$ by using the Pythagorean identity $$\cos^2\theta=1-\sin^2\theta$$. Let $$x=\sin\theta$$ for simplicity. Substituting $$\cos^2\theta=1-x^2$$ gives

$$2(1-x^2)-5x+4x^2=0.$$

Expanding and collecting like terms:

$$2-2x^2-5x+4x^2=0,$$

$$2+2x^2-5x=0.$$

Re-ordering, we get the quadratic equation

$$2x^2-5x+2=0.$$

To find its roots we use the quadratic formula. For $$ax^2+bx+c=0$$, the roots are $$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$$. Here $$a=2,\;b=-5,\;c=2$$, so

$$x=\dfrac{5\pm\sqrt{(-5)^2-4\cdot2\cdot2}}{2\cdot2} =\dfrac{5\pm\sqrt{25-16}}{4} =\dfrac{5\pm\sqrt{9}}{4} =\dfrac{5\pm3}{4}.$$

This gives two numerical roots:

$$x_1=\dfrac{5+3}{4}=2,\qquad x_2=\dfrac{5-3}{4}=\dfrac12.$$

Since $$x=\sin\theta$$ must satisfy $$-1\le x\le1$$, the value $$x=2$$ is impossible. Thus the only admissible solution is

$$\sin\theta=\dfrac12.$$

Within the interval $$(0,2\pi)$$ (excluding $$\pi$$) the equation $$\sin\theta=\dfrac12$$ is satisfied at

$$\theta=\dfrac\pi6\quad\text{and}\quad\theta=\dfrac{5\pi}6.$$

There are no other values in $$\bigl(\pi,2\pi\bigr)$$ with $$\sin\theta=\dfrac12$$ because $$\sin\theta$$ becomes negative beyond $$\pi$$. Therefore,

$$\theta_1=\dfrac{\pi}{6},\qquad \theta_2=\dfrac{5\pi}{6}.$$

Now we have to evaluate the definite integral

$$\int_{\theta_1}^{\theta_2}\cos^2 3\theta\,d\theta =\int_{\pi/6}^{5\pi/6}\cos^2 3\theta\,d\theta.$$

We use the power-reduction identity for the square of cosine:

$$\cos^2\alpha=\dfrac{1+\cos2\alpha}{2}.$$

With $$\alpha=3\theta$$, this becomes

$$\cos^2 3\theta=\dfrac{1+\cos6\theta}{2}.$$

Substituting into the integral gives

$$\int_{\pi/6}^{5\pi/6}\cos^2 3\theta\,d\theta =\int_{\pi/6}^{5\pi/6}\dfrac{1+\cos6\theta}{2}\,d\theta.$$

We can split and integrate term by term:

$$\int_{\pi/6}^{5\pi/6}\dfrac{1}{2}\,d\theta+\int_{\pi/6}^{5\pi/6}\dfrac{\cos6\theta}{2}\,d\theta =\dfrac12\int_{\pi/6}^{5\pi/6}d\theta+\dfrac12\int_{\pi/6}^{5\pi/6}\cos6\theta\,d\theta.$$

The first integral is straightforward:

$$\dfrac12\int_{\pi/6}^{5\pi/6}d\theta=\dfrac12\bigl[\theta\bigr]_{\pi/6}^{5\pi/6} =\dfrac12\left(\dfrac{5\pi}{6}-\dfrac{\pi}{6}\right) =\dfrac12\cdot\dfrac{4\pi}{6} =\dfrac{4\pi}{12} =\dfrac{\pi}{3}.$$

For the second integral we use the standard antiderivative $$\int\cos k\theta\,d\theta=\dfrac{\sin k\theta}{k}.$$ Here $$k=6,$$ so

$$\int\cos6\theta\,d\theta=\dfrac{\sin6\theta}{6}.$$

Therefore,

$$\dfrac12\int_{\pi/6}^{5\pi/6}\cos6\theta\,d\theta =\dfrac12\left[\dfrac{\sin6\theta}{6}\right]_{\pi/6}^{5\pi/6} =\dfrac{1}{12}\bigl[\sin6\theta\bigr]_{\pi/6}^{5\pi/6}.$$

Now we evaluate $$\sin6\theta$$ at the limits:

At $$\theta=\dfrac{5\pi}{6}:$$ $$6\theta=5\pi,$$ so $$\sin5\pi=0.$$

At $$\theta=\dfrac{\pi}{6}:$$ $$6\theta=\pi,$$ so $$\sin\pi=0.$$

Hence

$$\dfrac{1}{12}\bigl[\sin6\theta\bigr]_{\pi/6}^{5\pi/6} =\dfrac{1}{12}(0-0)=0.$$

Adding the two parts, the value of the integral is

$$\dfrac{\pi}{3}+0=\dfrac{\pi}{3}.$$

Hence, the correct answer is Option A.