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

In Exercises 11–24, find all solutions of each equation.
                 
4 sin θ﹣1 = 2 sin θ

Accepted Answer

Hello everybody. I hope you're 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 answers in radiance where our equation is nine multiplied by sign of Y is equal to 17, multiplied by sign of Y minus or multiplied by the square roots of three. Now, the answer choices provided are A Y equals pi divided by two plus pi multiplied by N where N is an integer B is Y is equal to pi divided by four plus pi N or pi is equal to or I'm sorry. Y is equal to three pi divided by four plus pi N where N is an integer C is Y is equal to pi divided by three plus two pi N or Y is equal to two pi divided by three plus two pi N or N is an integer N D Y is equal to pi divided by six plus two pi N or Y is equal to five pi divided by six plus two pi N where N is an integer. No, when we look at our equation, we're going to try to simplify it a bit before moving forward. So I see that I have sign of Y on the left and sign of Y on the right side of my equation. So we're trying to get them both on the same side. So we'll have four multiplied by the square of three. It's equal to 17 sign of why minus nine sign of Y. Now we'll have four multiplied by the squared of three is equal to 17, multiplied by the sign of Y minus nine sign of Y. We can just say 17 minus nine and we'll have eight sign of Y. And now if we try to isolate the sign of Y, we'll have that sign of Y is equal to four multiplied by the square of three and all that divided by eight, which can be simplified to have that sign of Y is equal to the square root of three divided by two. And I'm going to move my work a bit so that I can have more space to work with. And now let's recall from our unit circle values that sign of I divided by three is equal to the square root of three divided by two. Therefore, pine divided by three is a solution. Now to solve for other solutions, we're going to do a bit more work than just recall. Now, let's look at our side, our sign is positive. So where is sine positive? Well, sin is positive and two quadrants when you think of sign think of Y values. So where are Y values positive? We know that above the X axis, Y values are positive. So what quadrants lie above the X axis? Well, those will be quadrants one and two. Now we know that we already have pi divided by three as one of our solutions and pi divided by three is in quadrant one. Therefore, we are going to focus on quadrant two. Now for reference angles in order two, we have that reference, the reference angle is equal to pi minus the given angle. Now, in our case, we know our reference angle will be I divided by three and we're trying to solve for the given angle. So we have pi divided by three is equal to pi minus given angle. And once again, I'll move my work a bit. So we'll have three pi divided by three minus pi is equal to the angle. And finally, we'll have that the angle is equal to two pi divided by three because three pi divided by three minus pi is or we'll actually have three pie divided by three. That will be pie. And we'll have that minus pi divided by three, which was the reference angle that we had. So we'll have three pi divided by three minus pi divided by three. And that will leave us with two pi divided by three. So we have that our angle is two pi divided by three. So two pi divided by three will therefore be a solution. Now, let's think of the sine equation, the period of sign. And that is just how often in the graph, after what the period, after each period, the graph will continue to repeat. Now, the period of sign is two pi therefore, we have two solutions. Y is equal to pi divided by three, we said was a solution. But if we continue, let's say to an infinite graph, we'll have pi divided by three plus two pi N because N times two pi will show us that the graph will repeat every period. So that will be one of the solutions. And we'll also have that Y is equal to two pi divided by three plus two pi N where and is and integer. So that will be the solutions to this question. And we can see that those solutions match up with answer choice C so we can see how important it is to be able to recall different values from our unit circle as well as know different facts about the sine equation like the period and where it's going to be positive. So I hope that was helpful. And until next time.

In Exercises 11–24, find all solutions of each equation. 4 sin θ﹣1 = 2 sin θ

Key Concepts

Trigonometric Functions

Algebraic Manipulation

General Solutions of Trigonometric Equations

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