Can anyone help solve 2sinθ+1= cscθ

help would be GREATLY appreciated!

As cscθ = 1/sinθ, then by multiplying through by sinθ, we can eliminate cosec.

2sin^2θ + sinθ = 1

2sin^2θ + sinθ - 1 = 0

We now have a simple quadratic. Replace sinθ with y, to make it more manageable.

2y^2 + y - 1 = 0

(2y - 1)(y + 1)
y = 1/2, y = -1

sinθ = 1/2, sinθ = -1

Now, in the range 0 ≤ θ ≤ 360

For sinθ = 1/2,

θ = 30, 150

For sinθ = -1

θ = 270

So the values of θ that this works for are

θ = 30, 150, 270 (all in degrees; to change to radians, multiply each term by π/180)

Hope this helps :)

2sinθ+1= cscθ

2sinθ+1=1/sinθ

2sin^2 θ+1=1

2sin^2 θ

I'm not sure if I did this all correctly....it's definitely not finished....I tried my best.

I actually have a test on this tomorrow >___<

Must study more...