Question

For questions 3 and 4, what values for θ(0≤θ≤2π) satisfy the equation?

Answer 1

For that 1st one: divide both sides by sin (theta):

3 = 1 - 1/sin(theta)

1/sin(theta) = -2

sin(theta) = -1/2

Answer 2

Ok!

Answer 3

For that second one:

Multiply both sides by cos theta and then substitute 1- cos^2 for sin^2:

sin^2(theta) / cos (theta) = (-3/2)

1 - cos^2(theta) = (-3/2)cos(theta)

Now you have a quadratic in cos

Answer 4

I'm still not sure how to get the answers.

Answer 5

x^2 - (3/2)x - 1 = 0

2x^2 - 3x - 2 = 0 where x = cos(theta)

Answer 6

(2x + 1)(x - 2) = 0

The only one that makes sense for your roots is the first

2x + 1 = 0 -> x = -1/2 -> cos(theta) = -1/2

Answer 7

So, the last choice for b

Answer 8

Are you saying number 4 is D?

Answer 9

yes.

Answer 10

So is #3

Answer 11

Thank you so much!

Answer 12

uw! Bazinga!