Solve each equation for 0<_ 0<_ 2pi(3.14)
a) 2tan^20 = 3tan0 - 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`
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
thank you! i see where i did my calculations wrong..
To solve the equation, we need to find the value of θ (theta) that satisfies the equation for the given conditions, which are 0 ≤ θ ≤ 2π (3.14).
Let's start by simplifying the equation:
2tan^2(θ) = 3tan(θ) - 1
Next, let's substitute tan(θ) with its equivalent form using sin(θ) and cos(θ). Remember that tan(θ) = sin(θ) / cos(θ):
2(sin^2(θ) / cos^2(θ)) = 3(sin(θ) / cos(θ)) - 1
To get rid of the fractions, we can multiply the entire equation by cos^2(θ) (as long as cos(θ) ≠ 0, which we will verify later).
2sin^2(θ) = 3sin(θ)cos(θ) - cos^2(θ)
We can further simplify by using the trigonometric identity sin^2(θ) = 1 - cos^2(θ):
2(1 - cos^2(θ)) = 3sin(θ)cos(θ) - cos^2(θ)
Expanding and rearranging the terms, we get:
2 - 2cos^2(θ) = 3sin(θ)cos(θ) - cos^2(θ)
Combine like terms and move all terms to one side of the equation:
2 - cos^2(θ) - 3sin(θ)cos(θ) = 0
Now, we have a quadratic trinomial in terms of cos(θ). We can solve this by factoring or using the quadratic formula, but first, let's check if cos(θ) = 0 is a solution.
Since cos(θ) ≠ 0 (based on the given conditions 0 ≤ θ ≤ 2π), we can divide the entire equation by cos(θ):
(2 - cos(θ) - 3sin(θ))cos(θ) = 0
Now, we have two possibilities:
1. 2 - cos(θ) - 3sin(θ) = 0
2. cos(θ) = 0
For the first possibility, 2 - cos(θ) - 3sin(θ) = 0, we can solve it independently of the second possibility.
To simplify the equation further, let's rearrange the terms:
2 - cos(θ) = 3sin(θ)
Now, we can square both sides to eliminate the sin(θ) term:
(2 - cos(θ))^2 = (3sin(θ))^2
Expanding both sides, we get:
4 - 4cos(θ) + cos^2(θ) = 9sin^2(θ)
Using the trigonometric identity sin^2(θ) = 1 - cos^2(θ), we can substitute and simplify:
4 - 4cos(θ) + cos^2(θ) = 9(1 - cos^2(θ))
Expanding further:
4 - 4cos(θ) + cos^2(θ) = 9 - 9cos^2(θ)
Rearranging the terms:
10cos^2(θ) - 4cos(θ) - 5 = 0
Now, we have a quadratic equation in terms of cos(θ). We can solve it by factoring or using the quadratic formula. Let's use the quadratic formula:
cos(θ) = (-b ± √(b^2 - 4ac)) / (2a)
Substituting the values a = 10, b = -4, and c = -5:
cos(θ) = (-(-4) ± √((-4)^2 - 4(10)(-5))) / (2(10))
Simplifying:
cos(θ) = (4 ± √(16 + 200)) / 20
cos(θ) = (4 ± √(216)) / 20
cos(θ) = (4 ± 6√6) / 20
Simplifying further:
cos(θ) = (2 ± 3√6) / 10
cos(θ) = (1 ± (3√6) / 5
Now, we have two possible values for cos(θ):
1. cos(θ) = (1 + (3√6)) / 5
2. cos(θ) = (1 - (3√6)) / 5
To find the corresponding values of θ for the given conditions (0 ≤ θ ≤ 2π), we can use the inverse cosine function (cos^(-1)).
1. For cos(θ) = (1 + (3√6)) / 5:
θ = cos^(-1)((1 + (3√6)) / 5)
2. For cos(θ) = (1 - (3√6)) / 5:
θ = cos^(-1)((1 - (3√6)) / 5)
These are the solutions to the equation 2tan^2(θ) = 3tan(θ) - 1 for the given conditions 0 ≤ θ ≤ 2π. To obtain numerical values, you can use a calculator to evaluate the inverse cosine function.