Solve tan2A - 2tanA = -1 for 0º≤A≤360º.
remember your double-angle formula
tan2A = 2tanA/(1-tan^2 A)
so,
2tanA/(1-tan^2 A) - 2tanA = -1
2tanA * tan^2 A = tan^2 A - 1
2tan^3 A - tan^2 A + 1 = 0
Hmmm. Nasty. You need to solve
2x^3 - x^2 + 1 = 0
and then take the arctan of whatever solution you find.
To solve the equation tan2A - 2tanA = -1 for 0º ≤ A ≤ 360º, we'll use the substitution method. Let's substitute tanA with x:
tanA = x
Now, we need to express tan(2A) in terms of x:
tan(2A) = 2tanA / (1 - tan^2A)
= 2x / (1 - x^2)
Now, we can rewrite the equation as:
2x / (1 - x^2) - 2x = -1
To simplify, let's multiply both sides of the equation by (1 - x^2):
2x - 2x(1 - x^2) = -1(1 - x^2)
Simplifying further:
2x - 2x + 2x^3 = -1 + x^2
2x^3 - x^2 + 1 = 0
Now, we need to solve this cubic equation. There are various methods to solve cubic equations, such as factoring, the Rational Root Theorem, or using a graphing calculator. In this case, let's use trial and error to determine the possible solutions.
By examining the equation, we can see that there are no linear factors. Therefore, we'll assume that there is at least one rational root. Let's test some possible roots using synthetic division or a calculator until we find a factor that gives us a remainder of zero.
After attempting some values for x, we find that x = -1 is a root of the equation. Dividing the polynomial by (x + 1), we get:
(2x^3 - x^2 + 1) ÷ (x + 1) = 2x^2 - 3x + 1
Now, we can factor the quadratic expression:
2x^2 - 3x + 1 = (2x - 1)(x - 1)
Setting each factor equal to zero, we have:
2x - 1 = 0 --> x = 1/2
x - 1 = 0 --> x = 1
Since tanA = x, we have two values for x:
x = 1/2 and x = 1
Now, we need to find the corresponding values of A by substituting x back into the original equation tanA = x.
For x = 1/2:
tanA = 1/2
Taking the inverse tangent of both sides, we find:
A = tan^(-1)(1/2)
Using a calculator, we get:
A = 26.565º or A = 206.565º
For x = 1:
tanA = 1
Taking the inverse tangent of both sides, we find:
A = tan^(-1)(1)
Using a calculator, we get:
A = 45º or A = 225º
Therefore, the solutions for 0º ≤ A ≤ 360º are:
A = 26.565º, 45º, 206.565º, 225º