Solve each equation for 0<_ 0<_ 2pi(3.14)

Question

a) 2tan^20 = 3tan0 - 1

Answer 1

Let tan(θ) = x
So that,
2tan^2 (θ) = 3tan(θ) - 1
becomes
2x^2 = 3x - 1

Solving,
2x^2 - 3x + 1 = 0
(2x - 1)(x - 1) = 0
2x = 1
x = 1/2 : first root

x = 1 : second root

Substitute back tan(θ):
tan(θ) = 1/2
θ = tan^-1 (1/2)
θ = 26.6 degrees

tan(θ) = 1
θ = tan^-1 (1)
θ = 45 degrees

hope this helps~ `u`

Answer 2

There are actually 4 solutions,

from Jai's:

tanØ = 1/2
so by the CAST rule, Ø is in either the first or the third quadrant
Ø = 26.6° ---- Jai's answer
Ø = 180+26.6= 206.6°

if tanØ = 1
Ø = 45° , ----- again, Jai's answer
Ø = 180+45 = 225°

in degrees:
Ø = 26.6° , 206.6° , 45° , 225°

Furthermore, since the domain was stated as
0 ≤ Ø ≤ 2π , that is, in radians, we should state our answer in radians

Ø = .46365 , 3.60524 , π/4 , 5π/4

Answer 3

thank you! i see where i did my calculations wrong..