Question

What is the range of the values of k such that
k Cos A - 3 Sin A = k + 1

has a real solution?

Answer 1

k(cos A) - 3(sin A) = k + 1
k(cos A) - (k + 1) = 3(sin A)
[k(cos A) - (k + 1)]² = [3(sin A)]²
=> k²cos²A - 2k(k + 1)(cos A) + (k + 1)² = 9sin²A
=> k²cos²A - (2k² + 2k)(cos A) + (k + 1)² = 9(1 - cos²A)
=> (k² + 9)cos²A - (2k² + 2k)(cos A) + (k² + 2k - 8) = 0
Here we have a quadratic equation in terms of (cos A).
This equation is only solvable when the determinant b² - 4ac≥ 0, otherwise the solution will be complex.
b² - 4ac ≥ 0
=> [-(2k² + 2k)]² - 4(k² + 9)(k² + 2k - 8) ≥ 0
=> (4k^4 + 8k³ + 4k²) - (4k^4 + 8k³ + 4k² + 72k - 288) ≥ 0
=> -72k + 288 ≥ 0
=> 72k ≤ 288
=> k ≤ 4