Show that the equation 5sin(x) = 1 + 2 [cos(x)]^2 can be written in the form 2[sin(x)]^2 + 5 sin(x)-3=0

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Show that the equation 5sin(x) = 1 + 2 [cos(x)]^2 can be written in the form 2[sin(x)]^2 + 5 sin(x)-3=0

First, we need to realise that we will be using the trigonometric identity sin(x)² + cos(x)² = 1

As our goal is to end up with an equation involving only sin, we will therefore substitue cos(x)² with ( 1 - sin(x)²), giving

5sin(x) = 1 + 2(1-sin(x)²)

We then expand the brackets, getting

5sin(x) = 1 + 2 - 2sin(x)²

We want the final equation to equal 0, so we add make 1+2 equal 3 and subtract it from both sides of the equation:

5sin(x) -3 = -2sin(x)²

we then add 2sin(x)² on both sides, achieving the wanted equation:

2sin(x)²+ 5sin(x) -3 =0

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Show that the equation 5sin(x) = 1 + 2 [cos(x)]^2 can be written in the form 2[sin(x)]^2 + 5 sin(x)-3=0

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