Question

Let \(S = \{θ ∈ (0, 2π) : 7 cos^2θ – 3 sin^2θ – 2 cos^22θ = 2\}\). 
Then, the sum of roots of all the equations \(x^2 – 2 (tan^2θ + cot^2θ) x + 6 sin^2θ = 0, θ ∈ S\), is _______.

Answer 1

Correct Answer: 16

First, analyze the equation \(7 \cos^2θ - 3 \sin^2θ - 2 \cos^2(2θ) = 2\). Use trigonometric identities: \(\cos(2θ) = 2\cos^2θ - 1\) and \(\sin^2θ = 1 - \cos^2θ\). Since \(\cos^2(2θ) = \frac{1 + \cos(4θ)}{2}\), apply \(\cos(4θ) = 2\cos^2(2θ) - 1\) to simplify:

• \(2\cos^2(2θ) = 1 + \cos(4θ)\)

Substituting and simplifying:

• \(z = \cos^2θ, \sin^2θ = 1 - z\)
• \(7z - 3(1 - z) - (1 + 2z(2z - 1)) = 2\)
• \(7z - 3 + 3z - 1 - (1 + 4z^2 - 2z) = 2\)
• \(10z - 4z^2 - 4 = 2\)
• \(4z^2 - 10z + 6 = 0\)

Solve the quadratic \(2z^2 - 5z + 3 = 0\). Roots are \[z = \frac{1}{2}, z = \frac{3}{2}\]. But \(z = \cos^2θ ≤ 1\), so valid \(\cos^2θ = \frac{1}{2}\). Corresponding angle θ is:

• \(\cos θ = \pm \frac{1}{\sqrt{2}} ⇒ θ = \frac{π}{4}, \frac{3π}{4}, \frac{5π}{4}, \frac{7π}{4}\)

Use these θ to solve \(x^2 - 2(\tan^2θ + \cot^2θ)x + 6\sin^2θ = 0\):

• \(tan θ = \pm 1, tan^2θ = 1, cot^2θ = 1\)
• \(x^2 - 2(2)x + 6(1/2) = 0\)
• \(x^2 - 4x + 3 = 0\)
• Roots: \(x = 3, 1\). Sum: 4\)

So sum over all equations is \(4 × 4 = 16\). The final sum of roots is 16, verifying the provided range (16,16).

Answer 2

7 cos²θ – 3 sin²θ – 2 cos²2θ = 2
\(\begin{array}{l} \Rightarrow 4\left(\frac{1+\cos2\theta}{2}\right)+3\cos2\theta-2\cos^22\theta=2 \end{array}\)
⇒ 2 + 5 cos²θ – 2 cos² 2θ = 2
⇒ cos 2θ = 0 or 5/2 (rejected)
\(\begin{array}{l} \Rightarrow \cos2\theta=0=\frac{1-\tan^2\theta}{1+\tan^2\theta}\Rightarrow \tan^2\theta =1\end{array}\)
∴ Sum of roots = 2 (tan²θ + cot²θ) = 2 × 2 = 4
But as tanθ = ±1 for \(π/4, 3π/4, 5π/4, 7π/4\) in the interval (\(0, 2π\))
∴ Four equations will be formed
Hence, sum of roots of all the equations = 4 × 4 = 16.