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:

(A) $\frac{\pi}{3}+\frac{1}{6}$

Trigonometric Equations — JEE Main · 2020 · 07 Jan 2020 · Shift 2 Question with Solution

θ_{1}

and

θ_{2}

be respectively the smallest and the largest values of

θ

(0, 2 π) - {π}

which satisfy the equation,

2 cot^{2} θ - \frac{5}{s i n θ} + 4 = 0

, then

\int_{θ_{1}}^{θ_{2}} cos^{2} 3 θ d θ

is equal to:

Answer: (B)

Writing

cot^{2} θ = \frac{c o s ^{2} θ}{s i n ^{2} θ}

and clearing the denominator,

2 cos^{2} θ - 5 sin θ + 4 sin^{2} θ = 0

. Replacing

cos^{2} θ

with

1 - sin^{2} θ

gives

2 sin^{2} θ - 5 sin θ + 2 = 0

, i.e.

(2 sin θ - 1) (sin θ - 2) = 0

. Only

sin θ = \frac{1}{2}

is possible, so the smallest and largest solutions are

θ_{1} = \frac{π}{6}

and

θ_{2} = \frac{5 π}{6}

. Then

\int_{π /6}^{5 π /6} cos^{2} 3 θ d θ = \int_{π /6}^{5 π /6} \frac{1 + c o s 6 θ}{2} d θ = \frac{1}{2} [θ + \frac{s i n 6 θ}{6}]_{π /6}^{5 π /6} = \frac{1}{2} (\frac{5 π}{6} - \frac{π}{6}) = \frac{π}{3}

JEE Main · 2020 · 07 Jan 2020 · Shift 2 · Single correct option