how do you figure out a trigonometric equation like this one below?

Question

2cos^3+5cos^2+cos-2=0, where 0<0,360

do you find quadrant 3 and 4 for all 3 answers?

Answer 1

2cos³θ + 5cos²θ + cosθ - 2 = 0, where 0 ≤ θ < 360

Factor this just like you would

2x³ + 5x² + x - 2 = 0
(x + 2)(x + 1)(2x - 1) = 0

(cosθ + 2)(cosθ + 1)(2cosθ - 1) = 0
cosθ = -2, -1, 1/2

The first solution can be rejected because | cosθ | ≤ 1

cosθ = -1, 1/2
θ = 60°, 180°, 300°

Answer 2

HINT:

Replace cosine with any letter of choice and solve it like a regular quadratic equation.

We have:

2cos^3+5cos^2+cos-2=0

I will use x for cos and it looks like this:

2x^3 + 5x^2 + x - 2 = 0

Set into two groups.

2x^3 + 5x^2 = group 1

and

x - 2 = group 2.

We can factor 2x^3 + 5x^2 to be x^2(2x + 5)

We now have three factors:

x^2 + 1, 2x + 5 and x - 2.

Replace x with cos and set to zero to solve for cosine.

We now have:

cos^2 + 1 = 0

2cos + 5 = 0

cos - 2 = 0

Can you take it from here?

Guido

Answer 3

First of all, make a substitution in your mind t=cos. You will have a polynomial. To solve that you need to recall one rule - in this type of polynomial equations one of the solutions is a divisor of (free member)/(coefficient in front of the highest term). In this case this is 2/2=+ or -1. You substitute and find out that -1 is a solution. you then proceed to factorize the equation into this form (x-x1)(x-x2)(x-x3)=0 where xi - solutions.After you have all 3 solutions, you solve the equation cos=xi for each one.