1. ## Solve equation

Solve the equation over the intervals [0,2pi)

3cosx + 3 = 2sin²x

2sin²x - 3cosx - 3 = 0 Put it all on one side

1-2cos²x - 3cosx - 3 = 0 Used the pythagorean identity that Sin²x + Cos²x = 1

-2cos²x - 3cosx - 2 = 0 Combined the integers

From here i was confused because it doesn't factor into two binomials due to -cos²x being negative. So.. i made -cos²x positive using the even/odd identity. I am not sure if that is right.

(2cosx + 1)(cosx - 2) = 0 Used even/odd identity

So cosx = -1/2, 2

2pi/3, 4pi/3 My answer, didn't know what to do with the 2

Is this correct?

2. Originally Posted by brandonb122
Solve the equation over the intervals [0,2pi)

3cosx + 3 = 2sin²x

2sin²x - 3cosx - 3 = 0 Put it all on one side

2-2cos²x - 3cosx - 3 = 0 Used the pythagorean identity that Sin²x + Cos²x = 1

-2cos²x - 3cosx - 2 = 0 Combined the integers

From here i was confused because it doesn't factor into two binomials due to -cos²x being negative. So.. i made -cos²x positive using the even/odd identity. I am not sure if that is right.

(2cosx + 1)(cosx - 2) = 0 Used even/odd identity

So cosx = -1/2, 2

2pi/3, 4pi/3 My answer, didn't know what to do with the 2

Is this correct?
See my note in blue. You need to expand the 2 across the 1 as well as cos^2(x)

$\displaystyle -2\cos^2(x)-3\cos(x) -1 = 0 \implies 2\cos^2(x) + 3\cos(x)+1 = 0$

edit: $\displaystyle b^2-4ac = 1$ so this factors

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# 2sinĀ²X 3cosX - 3 = 0

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