I need help with a trig problem

Question

tan^3(2x-pi/6)+sec^2(2x-pi/6)-3tan(2x-pi/6)-4=0

Answer 1

This is one of those problems that's a little contrived but still, it gives you a chance to work on your trig identities and eventually division of polynomials.

tan^3(2x-pi/6)+sec^2(2x-pi/6)-3tan(2x-pi/6)-4=0

First you see that equation is (for the most part) a polynomial in the tangent of (2x-π/6), BUT you have the square of the secant of the same angle. To avoid carpal tunnel syndrome, I will, for the time being use θ to represent 2x - π/6.

now secθ is equal to 1/cosθ

if you square secθ that would be

sec²θ = 1/cos²θ

now if you use the identity cos²θ + sin²θ = 1

you can rewrite it further as

sec²θ = 1/cos²θ = (cos²θ + sin²θ)/cos²θ

sec²θ = 1 + (sin²θ/cos²θ) = 1 + tan²θ

and if we put the rearranged term into the original equation we have

tan³θ + (tan²θ+1) - 3tanθ -4 =0

tan³θ + tan²θ- 3tanθ -3 =0

So now we see this as a cubic polynomial of tanθ

What do we do from here?

we first find the roots of the polynomial; first we set y = tanθ

y³ + y² -3y -3 = 0

Now in general, the roots for such an equation will be difficult to find, but (here's the contrived part), this one can be solved using polynomial division. If it's one of those equations that come out with whole number roots, since the last term is -3, we have the possibility of either -1 and 3 or 1 and -3.

You can try all of them, but lets use -1 so we divide the equation by y+1

y+1 | y³ + y² -3y -3

y²

y+1 | y³ + y² -3y -3

1. (y³ + y²)

y² -3

y+1 | -3y - 3

-(-3y-3)

so y³ + y² -3y -3 = (y+1)(y² -3) = (y+1)(y - sqrt(3)) (y + sqrt(3)) = 0

so the roots of y are -1, and ±sqrt(3)

tanθ = -1 θ = 3π/4 (corresponding to 135 degrees) or 7π/4 (corresponding to 315 degrees)

θ = 2x-π/6 = 3π/4 -> 2x = 11π/12 , x = 11π/24

θ = 2x-π/6 = 7π/4 -> 2x = 23π/12 , x = 23π/24

tanθ = +sqrt(3) -> θ = π/3 (corresponding to 60 degrees) and θ = 4π/3 (corresponding to 240 degrees)

θ = 2x-π/6 = π/3 -> 2x = π/2 -> x = π/4

θ = 2x-π/6 = 4π/3 -> 2x = 3π/2 -> x = 3π/4

tanθ = -sqrt(3) -> θ = 2π/3 (corresponding to 120 degrees) and θ = 5π/3 (corresponding to 300 degrees)

θ = 2x-π/6 = 2π/3 -> 2x = 5π/6 -> x = 5π/12

θ = 2x-π/6 = 5π/3 -> 2x = 11π/6 -> x = 11π/12

(Again, I call upon my mathematical brothers and sisters to check my work)