Why do we add intervals of pi halves to our solution

The video tutorial explains how to find all solutions for the equation tangent equals ±1. It begins by identifying solutions within the interval 0 to 2π using the unit circle. The instructor then demonstrates how to graph the tangent function to visualize solutions beyond this interval. By understanding the periodic nature of the tangent function, the instructor shows how to extend solutions using π halves increments, avoiding redundancy. The final solution is expressed as X = π/4 + π/2 N, covering all possible solutions.

What are the angles where tangent equals plus or minus one within the interval of 0 to 2π?

• π/6, π/3, 2π/3, 5π/6

• π/4, 3π/4, 5π/4, 7π/4

• π/2, π, 3π/2, 2π

• π/3, 2π/3, 4π/3, 5π/3

What is the significance of vertical asymptotes in the tangent function graph?

• They represent the x-intercepts of the function.

• They show where the function has maximum values.

• They mark the points of inflection.

• They indicate where the function is undefined.

How can you find additional solutions for tangent equals plus or minus one beyond the interval of 0 to 2π?

• By subtracting π from each solution.

• By adding π/2 to each solution.

• By adding 2π to each solution.

• By multiplying each solution by 2.

What is the issue with simply adding 2π to each solution to find all solutions?

• It does not cover all possible solutions.

• It only works for positive angles.

• It leads to redundant solutions.

• It results in solutions that are too large.

What is the more efficient method for finding all solutions for tangent equals plus or minus one?

• Adding π/4 to each solution.

• Adding π/2 to each solution.

• Adding π to each solution.

• Adding 3π/4 to each solution.

What is the final formula for all solutions of tangent equals plus or minus one?

• X = π/4 + π/2n

• X = π/4 + 2πn

• X = π/2 + πn

• X = π/4 + πn

Why is it unnecessary to write all separate solutions when using the efficient method?

• Because it simplifies the calculation.

• Because it avoids redundancy.

• Because it is a faster method.

• Because it only works for positive angles.