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
SG
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