Let S={θ ∈(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 x²-2(tan ² θ+cot ² θ) x+6 sin ² θ=0, θ ∈ S, is __________.

Solution

Maths sir · 4 steps

Rewrite everything in cos 2θ, and the whole equation collapses to cos 2θ = 0.

1. One variable only. Put c = cos 2θ, so cos² θ = (1+c)/2 and sin² θ = (1-c)/2. Then 7 cos² θ - 3 sin² θ = (7+7c-3+3c)/2 = 2 + 5c.

   2 + 5c - 2c² = 2 ⇒ c (5 - 2c) = 0

2. Find the set S. c = 5/2 is impossible, so cos 2θ = 0. For θ ∈ (0, 2π) this gives θ = π/4, 3π/4, 5π/4, 7π/4, which is four values.
3. Build each quadratic. At each of these angles tan² θ = cot² θ = 1 and sin² θ = 1/2. So every equation is x² - 4x + 3 = 0, and its sum of roots is 4.
4. So. There is one equation for each of the 4 values of θ, and each contributes a sum of 4. The total is 4 × 4 = 16.

The common slip: 4. Counting the four identical equations as a single equation gives 4. The question adds the roots over every θ in S, and S has four elements.

The idea to keep. When cos² θ, sin² θ and cos² 2θ appear together, convert all of them to cos 2θ. You get a quadratic in one variable. Throw out any root that lies outside [-1, 1].

Answer from the NTA final answer key. Solution by PrepWiser.