In Exercises 11–24, find all solutions of each equation.

tan x =...

Question

In Exercises 11–24, find all solutions of each equation.

tan x = 1

Accepted Answer

Hello everybody. I hope you are doing. All right. Today, today we're going to be looking at this question that states for the following trigonometric equation solve and give the answer in radiance where the equation is tangent of B is equal to the square root of three. Now the answer choices provided are A B is equal to pi divided by three plus two pi N where N is an integer and it'll be an integer for all the answer choices I provide as your choice. B says that B is equal to pi divided by three plus pi and C says that B is equal to pi divided by six plus two pi N. And answer choice D says that B is equal to pi divided by six plus pi N. Now, the first thing I'm going to do here is do a simple recall of what tangent is exactly. So let's recall that tangent of being is equal to a sign of B divided by cosine of B. Now, we can continue to rec off from our unit circle. So let's say from the unit circle, let's consider an a specific angle. Let's consider pi divided by three. And this is something we should be able to recall unit circle values. We should be able to recall that. Now at pi divided by three, we have that sign of pi divided by three is equal to the square root of three divided by two and cosine of pi divided by three is equal to one half. Therefore, tangents of pi divided by three is equal to a square root of three divided by two. And the numerator divided by one half and the nominator. Now, in this fraction, the tools will cancel out and will be left with square roots of three. Therefore, tangent of pi over divided by three is squared of three, which means that pi divided by three is a solution. And now we have to see if there are other solutions to our question and to do that, let's move my work a bit to have more space. And let's see which quadrant we will be working in. Now, we know that tangent is positive because they give us that tenure of B is a positive squared of three. So where's tangent positive? Well, if tangent er of B is equal to sine of B divided by cosine of B for tangent to be positive, we need to have quadrants where both sine and cosine are the same sign and that occurs in quadrants one where they are both positive and three where they are both negative. Now, for quadrant one, we already have an angle or a solution pi divided by three. So we're going to be looking for a solution in quadrant three. Now, for the other solutions or solution, we have, we know we're going to be working in quad and three. Now for this angle in quad at three, let's try to solve for its reference angle in quarter three, the reference angle, we have a rule that for an angle in quantity three, the reference angle is equal to the angle that we want minus pi. And our reference angle we already solved for it. We obtain that in our first solution pi divided by three is a reference angle. So we can have that pi divided by three is equal to the angle that we want for our second solution minus pi. So we have that pi divided by three plus pi is equal to the angle that we're looking for. So we'll have pi divided by three plus three. Pi divided by three means that four pi divided by three is equal to the angle that we're looking for. And our second solution. Now, before we go on, let's talk about the period of tangent and its graph. Now for a graph of tangents, we know that the graph of trigonometric expressions repeats over the, the graph and they repeat every period. Now, the period of is P which means that every pi units that tangent graph is going to repeat, it might have different values but the general shape of the graph will repeat and will give us another solution to our equation. Therefore, we can apply that to our solutions. Now, we had two solutions. So we have B equals pi divided by three was our first solution and our second solution was B equals four pi divided by three. Now, to take into account the fact that the graphs go on infinitely and we can have multiple solutions We add to each of these solutionss hi, which is the period multiplied by end. So it could be multiple periods. And that just means that every pie and or every sole number of periods will have another solution to our equation. And just like that those two will be the solution for our equation. Now, we can also consider the fact that four pi divided by three plus pi N can be ignored because Pi divided by three plus pi multiplied by one is four pi divided by three already. So just using our first answer, we can obtain our second solution. So we ignore the second solution and only solve for the first one which is B equals pi divided by three plus pi N which matches with answer choice B and can show us how important it is to be able to recall how the values of tangent work that uh recall different values from our unit circle and know how to calculate for different reference angles. So I hope that was helpful and until next time.

In Exercises 11–24, find all solutions of each equation. tan x = 1

Key Concepts

Tangent Function

Inverse Trigonometric Functions

General Solution of Trigonometric Equations

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