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Solution :
Given, 4 cosθ - 3 secθ = tanθ
⇒ 4 cosθ - 3/cosθ = sinθ/cosθ
⇒ (4 cos²θ - 3)/cosθ = sinθ/cosθ
⇒ 4 cos²θ - 3 = sinθ [ cosθ ≠ 0 ]
⇒ 4 (1 - sin²θ) - 3 = sinθ
⇒ 4 - 4 sin²θ - 3 = sinθ
⇒ 4 sin²θ + sinθ - 1 = 0
∴ sinθ = {- 1 ± √(1 + 16)}/8
⇒ sinθ = (- 1 ± √17)/8
⇒ θ = sin⁻¹ (- 1 ± √17)/8
∴ the general solution is
θ = nπ + (- 1)ⁿ . sin⁻¹ (- 1 ± √17)/8 , n ∈ ℤ
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