Find all solutions between 0 and 2pi to tan(x)=-Squareroot3 exactly.

To find all solutions between 0 and 2π to the equation tan(x) = -√3, we can follow these steps:

1. Take the inverse tangent (arctan or tan^(-1)) of both sides of the equation to isolate x: arctan(tan(x)) = arctan(-√3).

This simplifies to: x = arctan(-√3).

2. Evaluate the arctan(-√3) using a calculator. This will give you the principal value (the value between -π/2 and π/2).

arctan(-√3) is approximately -1.0472 radians or -60 degrees.

3. However, we know that tangent is a periodic function with a period of π. So, we can add or subtract multiples of π to the principal value to get all solutions within the given range.

Let n be an integer. Then, the general solution is:
x = -1.0472 + nπ.

4. To find the solutions within the range of 0 to 2π, substitute n = 0, 1, 2, ..., 6 into the general solution.

x = -1.0472 + 0π = -1.0472 radians (or approximately -60 degrees).
x = -1.0472 + 1π = 2.0944 radians (or approximately 120 degrees).
x = -1.0472 + 2π = 5.2359 radians (or approximately 300 degrees).
x = -1.0472 + 3π = 8.3774 radians (or approximately 480 degrees).
x = -1.0472 + 4π = 11.5189 radians (or approximately 660 degrees).
x = -1.0472 + 5π = 14.6604 radians (or approximately 840 degrees).
x = -1.0472 + 6π = 17.8019 radians (or approximately 1020 degrees).

So, the solutions between 0 and 2π for tan(x) = -√3 are approximately:
x ≈ -1.0472, 2.0944, 5.2359, 8.3774, 11.5189, 14.6604, and 17.8019 radians (or approximately -60, 120, 300, 480, 660, 840, and 1020 degrees).

To find all solutions between 0 and 2π for the equation tan(x) = -√3, we can follow these steps:

Step 1: Identify the restricted domain of the tangent function:
The tangent function has vertical asymptotes at odd multiples of π/2. Therefore, to find the solutions within the given range, we need to find the values of x that are in between these vertical asymptotes.

Step 2: Find the principal solution:
The principal solution is the unique solution to the equation within the restricted domain. We can find it by taking the inverse tangent (arctan) of both sides of the equation.

arctan(tan(x)) = arctan(-√3)
x = arctan(-√3)

Using a calculator or reference table, we can find that arctan(-√3) ≈ -π/3.

So, the principal solution is x = -π/3.

Step 3: Find the general solutions:
To find the general solutions, we need to consider the periodic nature of the tangent function. The tangent function repeats every π radians or 180 degrees.

Since the principal solution is in the fourth quadrant (negative angle), we can add π to the principal solution to obtain another solution in the range of 0 to 2π.

x = -π/3 + π
x = 2π/3

Step 4: Determine if there are any additional solutions within the restricted domain:
We need to check if there are any other solutions in the given range between 0 and 2π.

If tan(x) = -√3, then tan(x) = √3 (by symmetry).
So, we need to find solutions to tan(x) = √3 within the restricted domain.

Using the same process as before, arctan(√3) ≈ π/3.

The general solution within the restricted domain is:
x = π/3 + π
x = 4π/3

Step 5: Finalize all solutions:
The solutions to the equation tan(x) = -√3 within the range of 0 to 2π are:
x = -π/3, 2π/3, 4π/3.

Therefore, the solutions are -π/3, 2π/3, and 4π/3.