Solve tan2 x + tan x – 1 = 0 for the principal value(s) to two
decimal places.
Please Explain
* Solve tan^2x + tanx-1=0 for the principal value(s) to two decimal places.
Let y=tan(x)
y²+y-1=0
use quadratic formula
y=(-1±√(5))/2
or
tan(x)=(-1±√(5))/2
find tan-1(-1±√(5))/2 using your calculator.
To solve the equation tan^2(x) + tan(x) - 1 = 0, we can use a substitution to simplify the equation. Let's say u = tan(x), then the equation becomes u^2 + u - 1 = 0.
Now, we can solve this quadratic equation for u. One way to do this is by factoring, but in this case, the equation does not factor nicely. So, we can use the quadratic formula: u = (-b ± √(b^2 - 4ac)) / (2a).
For our equation u^2 + u - 1 = 0, the coefficients of the equation are:
a = 1
b = 1
c = -1
Using the quadratic formula, we have:
u = (-1 ± √(1^2 - 4(1)(-1))) / (2 * 1)
u = (-1 ± √(1 + 4)) / 2
u = (-1 ± √5) / 2
So, we have two possible values for u: u = (-1 + √5) / 2 and u = (-1 - √5) / 2.
Now, we use the inverse tangent function to find the values of x associated with these values of u. Recall that tan(x) = u. Therefore:
x = arctan(u)
Using a calculator, we find:
For u = (-1 + √5) / 2:
x = arctan((-1 + √5) / 2) ≈ 0.786, rounded to two decimal places.
For u = (-1 - √5) / 2:
x = arctan((-1 - √5) / 2) ≈ -1.322, rounded to two decimal places.
So, the principal values of x to two decimal places that satisfy the equation tan^2(x) + tan(x) - 1 = 0 are approximately x = 0.79 and x ≈ -1.32.