Find all solutions to the equation with 0≤α≤2π. Give an exact answer if possible, otherwise give value(s) of α accurate to at least four decimal places.
4tan(α)+5=2a=
4tan(x)+5=2
tanx = -3/4
x will be in QII or QIV
To find all solutions to the equation 4tan(α) + 5 = 2a, we'll need to isolate the variable α. Let's begin by subtracting 5 from both sides:
4tan(α) = 2a - 5
Next, divide both sides by 4:
tan(α) = (2a - 5) / 4
Now, to find the solutions, we need to take the inverse tangent (or arctan) of both sides. However, since we are dealing with trigonometric functions, the solutions will repeat periodically, so we need to consider the unit circle and the range within 0 ≤ α ≤ 2π.
Let's find the general solution by taking the inverse tangent of both sides:
α = arctan((2a - 5) / 4)
The inverse tangent (or arctan) function returns a value in radians, so this will be the general solution. However, to get the exact solutions or specific values accurate to four decimal places, we'll need to plug in values for 'a'.
For each value of 'a', substitute it into the equation α = arctan((2a - 5) / 4), and compute the corresponding value(s) of α.
Keep in mind that to express the solutions within the range 0 ≤ α ≤ 2π, we may need to add or subtract multiples of 2π to the values of α obtained from the equation.
Note: Make sure to check the domain of 'a' so that the expression (2a - 5) / 4 is a valid input for the arctan function.
To find the solutions to the equation 4tan(α) + 5 = 2a, we can simplify and rearrange the equation as follows:
4tan(α) = 2a - 5
Now, divide both sides of the equation by 4:
tan(α) = (2a - 5) / 4
We know that tan(α) is equal to its corresponding sine and cosine values:
sin(α) / cos(α) = (2a - 5) / 4
Multiply both sides of the equation by cos(α):
sin(α) = ((2a - 5) / 4) * cos(α)
Now, we can square both sides of the equation:
sin^2(α) = ((2a - 5) / 4)^2 * cos^2(α)
Since sin^2(α) + cos^2(α) = 1, we can substitute (1 - cos^2(α)) for sin^2(α):
1 - cos^2(α) = ((2a - 5) / 4)^2 * cos^2(α)
Now, let's solve for cos(α):
cos^2(α) = 1 - ((2a - 5) / 4)^2 * cos^2(α)
Move the term involving cos^2(α) to the left side:
cos^2(α) + ((2a - 5) / 4)^2 * cos^2(α) = 1
Factor out cos^2(α):
(cos^2(α)) * (1 + ((2a - 5) / 4)^2) = 1
Now, we can solve for cos(α):
cos(α) = ±sqrt(1 / (1 + ((2a - 5) / 4)^2))
Finally, to find the solutions for α, we take the inverse cosine of both sides:
α = arccos(±sqrt(1 / (1 + ((2a - 5) / 4)^2)))
Note that the symbol ± indicates the solutions will have both positive and negative values. To find the exact solutions, plug in the values of a. If the calculations yield values of α that are not exact, round them to at least four decimal places for an approximate answer.