Verified answer

let y=sec(θ)

2y^2=3y+2

2y^2-3y-2=0

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

y=-1/2 or y=2

sec(θ)=-1/2 or sec(θ)=2

cos(θ)=1/2,

so θ=60, or 300 degrees.

Note that sec(θ)=-1/2

isn't true since that says

cos(θ)=-2 and we know cosine falls in between

-1 and 1, inclusive.

I changed theta to x

2 sec^2 x = 3 sec x +2

0 = 2 + 3 sec x - 2 sec^2 x

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

1 + 2 sec x = 0 and 2 - sec x = 0

2 sec x = -1 and 2 = sec x

sec x = -1/2 and sec x = 2

1/cos x = -1/2 and 1/cos x = 2

cos x = -2 and cos x = 1/2

From cos x = 1/2,

x = 60 and 300

To find x from cos x = -2, you have to put it in the calculator, but right now, I'm thinking it doesn't exist.

2sec^2(θ) = 3sec(θ) + 2

Arrange to a quadratic :

2sec^2(θ) - 3sec(θ) - 2 = 0

Factorise :

[2sec(θ) + 1][sec(θ) - 2] = 0

Therefore, either,

2sec(θ) + 1 = 0, which implies sec(θ) = -1/2, or, cos(θ) = -2 (impossible)

or,

sec(θ) = 2, which implies cos(θ) = 1/2.

θ = 360ºn ± cos^-1(1/2)

= 360ºn ± 60º

The following gives 0 ≤ θ ≤ 360º

n = 0 implies θ = 60º (using the positive sign)

n = 1 implies θ = 300º (using the negative sign)