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# How do you solve 2 cot²θ - 3 cscθ equals 0?

Updated: 4/28/2022

Wiki User

13y ago

You have the following:

2cot2θ - 3cscθ = 0

Start by adding "3cscθ" to both sides:

2cot2θ = 3cscθ

Next, put the equation in terms of sine and cosine. This will make it easier to work with.

2cos2θ/sin2θ = 3/sinθ

Multiply both sides by sinθ to simplify:

2cos2θ/sinθ = 3

Multiply both sides by sinθ again, so the sinθ is on the other side:

2cos2θ = 3sinθ

Using the Pythagorean trigonometric identity (cos2θ + sin2θ = 1), we can put the equation entirely in terms of sine:

2(1 - sin2θ) = 3sinθ

2 - 2sin2θ = 3sinθ

Add 2sin2θ and subtract 2 from both sides, so the terms are all on one side:

0 = 2sin2θ + 3sinθ - 2

Notice how the equation looks similar to a quadratic equation. We can factor the equation just like a quadratic equation, so we get:

0 = (2sinθ - 1)(sinθ + 2)

If either piece is equal to zero, the entire equation is equal to zero. All we have to do is solve each part seperately for zero.

2sinθ - 1 = 0

2sinθ = 1

sinθ = 1/2

θ = pi/3 or 2pi/3

--

sinθ + 2 = 0

sinθ = -2

The second piece has no solution. The only two answers within the interval [0, 2pi] are pi/3 and 2pi/3.

Wiki User

13y ago