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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:...

Answered by Henry S. • Maths tutor

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Write 9sin(x) + 12 cos(x) in the form Rsin(x+y) and hence solve 9sin(x) + 12 cos(x) = 3

9sin(x) + 12 cos(x) = Rsin(x+y) =R(sin(x)cos(y)+cos(x)sin(y))= (Rcos(y))sin(x) + (Rsin(y))cos(x)Therefore by matching the coefficientsRsin(y)=12, Rcos(y)=9 [1]SoRsin(y)/Rcos(y) = 12/9 = 4/3, therefore tan...

Answered by James H. • Maths tutor

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g(x) = e^(x-1) + x - 6 Show that the equation g(x) = 0 can be written as x = ln(6 - x) + 1, where x<6

0 = e^x-1+ x - 6 e^x-1= 6-x x-1 = ln (6-x) -> here we have taken the natural log of both sides, but it only shows on one side as the natural log of e is 1.x = ln (6-x) + 1Question ...

Answered by Sumrah N. • Maths tutor

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