find all solutions to 4tanθ + √3 = tanθ
4tanθ + √3 = tanθ
3tanØ = -√3
tanØ = -√3/3
Ø must be in II or IV
Ø = 150° or 330°
or
Ø = 5π/6 or 11π/6
period of tanØ = 90°
general solutions:
=(150+90k)° or 330° +90k°
do the same with the radian answer by adding πk/2 to each of the answers
To find all solutions to the equation 4tanθ + √3 = tanθ, we need to isolate the variable θ.
Let's start by moving the term with √3 to the other side of the equation:
4tanθ = tanθ - √3
Next, let's combine the tanθ terms on the right-hand side:
4tanθ - tanθ = -√3
(4 - 1)tanθ = -√3
3tanθ = -√3
Now, divide both sides of the equation by 3 to solve for tanθ:
tanθ = -√3/3
Since we're looking for all solutions, we need to consider the unit circle and the periodicity of the tangent function. The tangent function repeats its values every π radians (or 180 degrees).
The value -√3/3 corresponds to a special angle on the unit circle, which is -π/6 radians or -30 degrees (in the third quadrant).
Since the tangent function repeats every π radians, we can add any multiple of π to the solution:
θ = -π/6 + nπ, where n is an integer.
So, the general solution to the equation 4tanθ + √3 = tanθ is:
θ = -π/6 + nπ, where n is an integer.