Given that: 2tanθsinθ = 4 - 3cosθ , show that: 0 = cos²θ - 4cosθ + 2 .
Starting with: 2tanθsinθ = 4 - 3cosθ . We can rewrite tanθ in terms of sinθ and cosθ.We know: tanθ = sinθ ÷ cosθ .By substituting we get: 2(sinθ ÷ cosθ)sinθ = 4 - 3cosθ .Let's multiply out by cosθ to get: 2sin²θ = 4cosθ - 3 cos²θ .Remembering the trigonometric identity:sin²θ + cos²θ = 1 . We can find that: 2sin² = 2 - 2cos²θ .This is useful because when we substitute back into the original equation we can eliminate the 2sin²θ term.Hence: 2 - 2cos²θ = 4cosθ - 3 cos²θ .Finally rearranging we get: 0 = cos²θ - 4cosθ + 2 . Just what we wanted.
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