If 2cosθ + sinθ = 1 (θ ≠ π/2), then 7cosθ + 6sinθ is equal to:

2cosθ + sinθ = 1, θ ≠ π/2
2cosθ = 1 - sinθ
4cos²θ = (1 - sinθ)²
Let sinθ = x
4(1 - x²) = (1 - x)²
4 - 4x² = 1 - 2x + x²
5x² - 2x - 3 = 0
(x - 1)(5x + 3) = 0
x = 1 ⇒ sinθ = 1 ⇒ θ = π/2 but given θ ≠ π/2
∴ x = -3/5
sinθ = -3/5
∴ cosθ = √(1 - sin²θ) = √(1 - 9/25) = √(16/25) = 4/5
∴ 7cosθ + 6sinθ = 7 × (4/5) + 6 × (-3/5) = 28/5 - 18/5 = 10/5 = 2