What is the total number of solutions for the equation 3 tan^2 A + tan A – 2 = 0 in the interval. 0 ≤ A ≤ π?
you have two solutions for tanA, one positive, one negative.
So, there are two values for A, one in each QI, one in QII.
To find the total number of solutions for the given equation in the interval 0 ≤ A ≤ π, we can use the discriminant to determine the nature of the solutions.
Step 1: Identify the coefficients of the quadratic equation.
The given equation is: 3 tan^2 A + tan A – 2 = 0
Comparing it with the general form of a quadratic equation ax^2 + bx + c = 0, we have:
a = 3
b = 1
c = -2
Step 2: Calculate the discriminant.
The discriminant (D) of a quadratic equation is given by the formula:
D = b^2 - 4ac
Substituting the given values, we have:
D = (1)^2 - 4(3)(-2)
D = 1 + 24
D = 25
Step 3: Analyze the discriminant.
If the discriminant is positive (D > 0), then there will be two real solutions.
If the discriminant is zero (D = 0), then there will be one real solution.
If the discriminant is negative (D < 0), then there will be no real solutions.
In this case, the discriminant D = 25, which is positive. Therefore, there will be two real solutions for the equation 3 tan^2 A + tan A – 2 = 0 in the interval 0 ≤ A ≤ π.
To find the total number of solutions for the given equation in the interval 0 ≤ A ≤ π, we can use the properties of the tangent function and algebraic techniques.
Step 1: Analyze the equation
The given equation is a quadratic equation in terms of tan A. Let's rewrite it by substituting x = tan A:
3x^2 + x - 2 = 0
Step 2: Factor or use the quadratic formula
To find the solutions for the quadratic equation, we need to either factor it or use the quadratic formula.
Since it may not be easy to factorize the equation, let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
Here, a = 3, b = 1, c = -2.
x = (-1 ± √(1^2 - 4 * 3 * -2)) / (2 * 3)
x = (-1 ± √(1 + 24)) / 6
x = (-1 ± √25) / 6
x = (-1 ± 5) / 6
So, x = (4/6) or x = (-6/6).
Simplifying, we get x = 2/3 or x = -1.
Step 3: Find the values of A
Now, we need to find the values of A by substituting back tan A = x:
For x = 2/3:
tan A = 2/3
Taking the inverse tangent (aka arctan) of both sides, we have:
A = arctan(2/3)
For x = -1:
tan A = -1
Taking the inverse tangent of both sides, we have:
A = arctan(-1)
However, the interval of interest is 0 ≤ A ≤ π, so we need to restrict the solutions within this range.
Step 4: Restrict the solutions within the interval
For the equation tan A = 2/3, we can use the inverse tangent function to find the principal value of A within the given interval. The principal value of arctan(2/3) lies between 0 and π/2.
Thus, A = arctan(2/3) satisfies the condition 0 ≤ A ≤ π.
For the equation tan A = -1, the principal value of arctan(-1) lies between -π/4 and π/4, but this does not overlap with the given interval 0 ≤ A ≤ π. Therefore, there are no solutions for this equation in the given interval.
Step 5: Determine the total number of solutions
From the previous steps, we find that there is only one solution for the equation 3 tan^2 A + tan A – 2 = 0 in the interval 0 ≤ A ≤ π.
Therefore, the total number of solutions for the given equation within the specified interval is 1.