If and respectively are the numbers of positive and negative values of in the interval that satisfy the equation , then is equal to _______.
Show answer and solution
Answer: 25
Solution
Using product-to-sum: cos2θ·cos(θ/2) = (1/2)[cos(5θ/2) + cos(3θ/2)] and cos3θ·cos(9θ/2) = (1/2)[cos(15θ/2) + cos(3θ/2)].
So the equation reduces to cos(5θ/2) = cos(15θ/2).
This gives either 15θ/2 = 5θ/2 + 2kπ → 5θ = 2kπ → θ = 2kπ/5, or 15θ/2 = -5θ/2 + 2kπ → 10θ = 2kπ → θ = kπ/5.
Note θ = 2kπ/5 is a subset of θ = kπ/5, so all solutions are θ = kπ/5 for integer k.
In [-π, π]: θ = -5π/5, -4π/5, -3π/5, -2π/5, -π/5, 0, π/5, 2π/5, 3π/5, 4π/5, 5π/5 → k = -5,-4,-3,-2,-1,0,1,2,3,4,5 giving 11 values.
Positive values (θ > 0): k = 1,2,3,4,5 → m = 5.
Negative values (θ < 0): k = -1,-2,-3,-4,-5 → n = 5.
So the equation reduces to cos(5θ/2) = cos(15θ/2).
This gives either 15θ/2 = 5θ/2 + 2kπ → 5θ = 2kπ → θ = 2kπ/5, or 15θ/2 = -5θ/2 + 2kπ → 10θ = 2kπ → θ = kπ/5.
Note θ = 2kπ/5 is a subset of θ = kπ/5, so all solutions are θ = kπ/5 for integer k.
In [-π, π]: θ = -5π/5, -4π/5, -3π/5, -2π/5, -π/5, 0, π/5, 2π/5, 3π/5, 4π/5, 5π/5 → k = -5,-4,-3,-2,-1,0,1,2,3,4,5 giving 11 values.
Positive values (θ > 0): k = 1,2,3,4,5 → m = 5.
Negative values (θ < 0): k = -1,-2,-3,-4,-5 → n = 5.
[JEE Main 2023 · 25 Jan · Shift 2]