Let \(\mathrm{S}=\left\{\theta \in(0,2 \pi): 7 \cos ^{2} \theta-3 \sin ^{2} \theta-2\right. \left.\cos ^{2} 2 \theta=2\right\}\). Then, the sum of roots of all the equations \(x^{2}-2\left(\tan ^{2} \theta+\cot ^{2} \theta\right) x+6 \sin ^{2} \theta=0 \theta \in \mathrm{S}\), is \_\_\_\_ .

Trigonometric Equations — JEE Main · 2022 · 29 Jul 2022 · Shift 1 Question with Solution

Let

S = {θ \in (0, 2 π) : 7 cos^{2} θ - 3 sin^{2} θ - 2 cos^{2} 2 θ = 2}

. Then, the sum of roots of all the equations

x^{2} - 2 (tan^{2} θ + cot^{2} θ) x + 6 sin^{2} θ = 0 θ \in S

, is ____ .

Answer: 16

Simplify the given equation:

7 cos^{2} θ - 3 sin^{2} θ - 2 cos^{2} 2 θ = 2

4 cos^{2} θ + 3 cos 2 θ - 2 cos^{2} 2 θ = 2

2 (1 + cos 2 θ) + 3 cos 2 θ - 2 cos^{2} 2 θ = 2

2 cos^{2} 2 θ - 5 cos 2 θ = 0

cos 2 θ (2 cos 2 θ - 5) = 0

Since

2 cos 2 θ - 5 = 0

has no solution (

cos 2 θ \leq 1

), we have

cos 2 θ = 0

, giving

2 θ = \frac{π}{2}, \frac{3 π}{2}, \frac{5 π}{2}, \frac{7 π}{2}

, so

θ = \frac{π}{4}, \frac{3 π}{4}, \frac{5 π}{4}, \frac{7 π}{4}

.

For each of these values,

sin^{2} θ = cos^{2} θ = \frac{1}{2}

, so

tan^{2} θ = cot^{2} θ = 1

. The equation becomes:

x^{2} - 2 (1 + 1) x + 6 \cdot \frac{1}{2} = 0 \Rightarrow x^{2} - 4 x + 3 = 0

Sum of roots = 4 for each equation. Since there are 4 values of

θ

, total sum =

4 \times 4 = 16

JEE Main · 2022 · 29 Jul 2022 · Shift 1 · Integer answer