11. Inverse Trigonometric Functions and Basic Trig Equations

Linear Trigonometric Equations

11. Inverse Trigonometric Functions and Basic Trig Equations

Linear Trigonometric Equations - Video Tutorials & Practice Problems

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Introduction to Trig Equations

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Hey, everyone in working with trig functions, you've likely come across a statement such as this one, the sign of the equals one half. But this statement is actually more explicitly a trig equation which we're going to be diving deeper into. Now. Now, in working with the unit circle, you likely also know that there are multiple angles for which this is true. The sign is equal to one half, which leads us to the conclusion that trig equations have multiple solutionss. But how exactly can we go about finding all of these solutions? Well, here I'm going to walk you through exactly how to find all the solutions to trigger equations such as this one just using the unit circle and adding in one extra thing. So let's go ahead and get started here. Now, in our example here, we're asked to find all of the solutions to the equation within the interval 0 to 2 pi which is just one rotation around our unit circle. So here we want to find all of the angles for which the S is equal to one half. So coming to our unit circle over here, starting in that first quadrant I know that for my angle pi over six, the S is equal to one half. So that represents a solution to my equation here beta equals pi over six. Now continuing around my unit circle going into quadrant two, I know that for five pi over six, the S is also equal to one half. So this represents yet another solution to my trig equation. Now in quadrants three and four, I know that my sine values are going to be negative. So there are no other solutions in here. So Pi over six and five pi over six represent the solutions to my trig equation within the interval 0 to 2 pi. But often you're not just going to be limited to that interval 0 to 2 pi and you're actually going to be asked to find all of the solutions to the equation with no restrictions. So what exactly does that mean to find all of the solutions to a trigger equation? Well, for my trigger equation here, the s of theta equals one half, I already know that Pi over six and five pi over six represent two of the solutions to this equation. But if I were to continue another rotation around my unit circle and come back to what would now be 13 pi over six, the sign of 13 pi over six is also equal to one half. So that represents yet another solution to this equation. And then if I were to continue on for another, going further into that rotation and reach what would now be 17 pi over six. The sign of 17 pi over six is also equal to one half. So this represents yet another solution. And if I were to continue going around and around and around my unit circle, I would have an infinite number of solutions. So how do we account for all of those solutions? Will, you may be worried that it's going to be a bit tricky? But it's actually rather simple because all we're going to do is first find all of the solutions on our unit circle and then simply add two pi in to each of them where two pi represents a full rotation around my unit circle and in is just an integer that represents how many times you're going around it. Now, you might also see this written as two pi K depending on your professor or your textbook, but it means the same exact thing. So let's look at our sign of data equals one half and identify all of the solutions using this method. Now we first want to find all of those solutions on our unit circle, which we already did in our earlier example. So we know that on our unit circle, we have our two solutions of pi over six and five pi over six that make this statement true. The sign of it is equal to one half. Now that we have these solutions that are on our unit circle. All that's left to do is add two pi in to each of them. And we've now covered all of our bases. This represents all of the possible solutions to this equation. The sign of the equals one half. And we're done here. This is our final answer. Now, let's take a look at one other example here. Here, we're asked to find all of the solutions to the equation, the cosine of X equals negative one. Now, the first thing you might notice here is that we're dealing with an X rather than a theta. But that doesn't change anything. We're still trying to find the angles that make this equation true. So coming up to our unit circle, we first want to find all of our solutions that are already on there and then go from there. So we want to find all of the solutions for which our cosine is equal to negative one. So on our unit circle here, if I go around looking for my cosine values of negative one, there's actually only one angle for which this is true. And that's pi so my solution that is on my unit circle is X equals pi. Now all that's left to do is add two pi N to this pi and we're done, this represents all of the possible solutions to the equation. The cosine of X equals negative one. And we're good to go now that we know how to find all of the solutions to basic trig equations. Let's get some more practice together. Thanks for watching and let me know if you have any questions.

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Example 1

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Hey, everyone in this problem, we're asked to find all solutions to the equation. The tangent of the is equal to the square root of three. So remember the first thing we want to do here is find all the angles for which this is true on our unit circle. Then we can proceed with adding two PN to each of those solutions. So starting on my unit circle, I want to find all the values for which my tangent is route three. So in quadrant one, I know that for my pi over three, my tangent is equal to the square root of three. Now also knowing that in quadrant three, all of my tangent values are positive here. And I'm looking for positive square root of three, I know that from my angle four pi over three, my tangent is also equal to the square root of three. So here two answers that I get are pi over three and also four pi over three. Now we can just go ahead and add two pi N to each of these solutions in order to get all of the solutions to this equation. So if we take both of these add two pi in this represents all of the solutions to our equation. But something that you actually might notice here is that pi over three and four pi over three are exactly pi radiance away from each other. So if I were to just take pi over three and add pi into it, I would still get all of my solutions. So starting with pi over three, if I add one rotation of pi, if I add another one and then another one and then another one, I will still get all of my solutions. So while Pi over three plus two Pi N and four Pi over three plus two PN still represent all of my solutions. If I want to further simplify this, I can write this as Pi over three plus pi N and this represents the most simplified form. I love all of my solutions because when working with a tangent, if I have two angles for which this is true, they are actually always going to be pi radiance away from each other. So this doesn't happen for every single trig function or trig equation that you work with. But you may notice that if your angles are exactly pi radiance away from each other, you can actually simplify this further. So our final answer here in its most simplified form is Pi over three plus Pi in for the tangent of the being equal to the square root of three. Thanks for watching and I'll see you in the next one.

Hey, everyone in this problem, we're asked to find all of the solutions to the equation, the sign of the equals one half. But we want to express these solutions in degrees. So we're going to have to keep that in mind as we go about solving this. Now, here we want to start by finding all of the solutions that are on our unit circle. So we want to find our angles for which the sign is equal to one half. Now coming over here to my unit circle, I know that the sign of pi over six is one half. And then also over here in quadrant two, that's the other quadrant for which my S values are positive. I know that the sine of five pi over six will also be positive one half. So my solutions on my unit circle are theta equals pi over six and theta equals five pi over six. Now I'm gonna stop here and go ahead and convert these into degrees. Now these are radiant measures. So whenever we have radiant measures, we can simply multiply them by 180 degrees over pi in order to convert them to degrees. So let's go ahead and do that to both of these radiant angle measures. So I want to multiply this by a 180 over pi and then multiply five pi over six also by 180 over pi. Now for that pi over six times 180 over pi those pies are of course going to cancel, leaving me with 180 divided by six. Now this gives me a value of 30 degrees. Now for five pi over six, again, those pies are going to cancel. And I'm going to be left with five times 180/6, which gives me an angle measure of 150 degrees. Now, you might also just have these values memorized from looking at your unit circle. But remember that you can always calculate them by simply multiplying your radiant angle measure by 180 over pi. Now, from here, we have our angle measures in degrees. But remember we want to find all of the solutions here. So we wanna add multiple rotations to these. Now, typically we add two pi in but two PN is in radiance. So instead of two pi I wanna express a full rotation in degrees. So instead of two PN I am instead going to have 360 degrees times N to each of these to represent multiple rotations but in degrees. So here my final solutions are 30 degrees plus 360 N and 150 degrees plus 360 N. Thanks for watching and I'll see you in the next one.

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How to Solve Linear Trigonometric Equations

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Hey, everyone, we just learned how to find all of the solutions to a basic trig equation using the unit circle. But as you continue to work through problems, these equations are going to start to get a bit more complicated and you might come across something that looks like this or sign of the minus three equals one. Now, as these equations begin to look more complicated, it's only natural to think that they're going to become more complicated to solve as well. But that's actually not the case here at all because we can take this a more complicated trick equation and turn it right back into a basic trig equation that we already know how to solve the same exact way we would solve for X in a linear equation, something that you're probably already really good at. So here I'm going to walk you through step by step exactly how to solve these more complicated trick equations. And soon you'll be able to solve any linear trick equation that gets thrown your way. So let's go ahead and jump right in here. Now with our basic linear equations here, we can just use our basic algebraic operations like addition and division in order to solve this and isolate our variable. In this case, X to get our final answer. Now, in working with a linear trig equation, we can do the exact same thing. But instead of isolating a variable, we're instead going to be isolating our trig function as though it is a variable. So in our linear trig equation that we have here four s of the minus three equals one, we're going to treat our trig function the sine of the as though it is a variable. So this sign of data is just like X. So since this is the same exact equation as the one we had over here just now with a sign instead of an X, we can apply the exact same operations in order to isolate that trig function ending up with sine of theta equals one. Now we're back to a basic trig equation that we can solve by using the unit circle. And we would end up getting that our angle theta is equal to pi over two. So all that we're doing here is isolating our trig function and then applying what we already know. Now, not all linear trig equations will be this straightforward. So here I'm going to walk you through some steps that will work for any linear trig equation. Now, looking at the example that we have here, we're given negative two cosine of theta plus the square root of three equals zero. And we're asked to find all of the solutions of this, of this equation within 0 to 2 pi. So let's go ahead and get started with step one here, which is to isolate our trig function using our basic algebraic operations. Now, here are trig function is the cosine of theta. So we're going to be treating that cosine as though it is a variable and isolating it. So here, the first step that I want to take is subtracting the square root of three from both sides canceling it on that left side, leaving me with a negative two cosine of theta is equal to negative square root of three. Now to fully isolate this trig function cosine of the I can divide both sides by negative two, canceling that negative two leaving me with the cosine of theta is equal two, positive route 3/2 with those negatives having canceled each other out. So now I fully isolated my trig function and I've completed step number one. Now remember that trig equations actually have multiple solutions. So in order to continue on here, we want to go ahead and continue on with step two and find all of the solutions that are on our unit circle. So here we have the cosine of data is equal to route 3/2. And we want to find all of the angles for which that is true. So coming back down here to my unit circle I know that for my angle pi over six, the cosine is root 3/2 and continuing on around my unit circle. I also know that here in quadrant four at 11 pi over six, the cosine will be the square root of 3/2 as well. So my two solutions here on my unit circle are theta equals pi over six and theta equals 11 pi over six. So we have completed step two, we have found all of the solutions on our unit circle and we can continue on to step three. Now, step three tells us that if our domain is not restricted, we want to go ahead and add two pi to each solution. But what does that mean? Our domain is not restricted? Well, in our problem statement, we were told to find all solutions within 0 to 2 pi now 0 to 2 pi is just one rotation around my unit circle. So my domain is restricted and I've already found my answers that are within that interval. So I've already completed step three because my domain is actually restricted. So I don't have to worry about doing anything additional here. Now, for step four, we're just left to isolate the. Now in our trig function here, the is already by itself in that argument of cosine. So I don't have to do anything for step four, either the is already isolated and by itself. So step four is done. And I have my final answers here. Theta equals pi over six and theta equals 11 pi over six. So now that we know how to solve linear trigger equations, let's continue to get some practice together. Thanks for watching and I'll see you in the next one.

Everyone in this problem, we're asked to find all solutions to the equation four times the cosine of two, theta plus pi plus eight is equal to 12. Now, this might look a little bit complicated at first, but we're still going to work through this the same exact way using our steps. So of course, starting with step number one, we want to go ahead and isolate our trig function. Now, in this case, our trig function looks a little bit different. It's a little bit longer. It is this full cosine of two theta plus pi. So that's what we wanna isolate. Now, we can start by subtracting eight from both sides canceling over here and leaving me with four times the cosine of two, theta plus pi is equal to 12 minus eight, which is positive four. Now, from here, I can isolate this by dividing both sides by four, canceling on that left side, leaving me just with my trig function cosine of two, theta plus pi is equal to four divided by four, which is positive one. Now I've completed step number one, my trig function is isolated. It's all by itself and I can move on to step number two, find all of my solutions on the unit circle. Now, this might seem a little bit strange to you at first because we're not just dealing with theta, we're dealing with two theta plus pi, but it really doesn't matter at this point in our process. It's not gonna matter until later on in our steps because we're still trying to find a value for which the cosine is equal to this one. So two theta plus pi, what does that need to be? In order to get a cosine of one, we're working through this the same way. So coming to our unit circle here, where is my cosine going to be equal to one? Well, I know that for zero, my cosine is one and that's actually the only angle for which that is true. So in order to get a cosine value of one, that two theta plus pi needs to be equal to zero. So from here we have completed step number two, we found our solutions that are on our unit circle. Now, from here, if our domain is not restricted, we want to add two pi N to each solution. And because we're asked to find all of our solutions and not given a restriction, we do need to go ahead and add two pi in over here. So here I'm left with zero plus two pi N. Now I can actually simplify that because that zero isn't doing anything. So this is two pi N on that right side and then I'm left with two, theta plus pi is equal to two pi N that represents all of my solutions. But we have one final step here because we've completed step number three. But step number four is to isolate the, that's where we're actually going to deal with this two theta plus pi. So from here, we're gonna go, just go ahead and use algebra to isolate the. So in order to do that, I want this theta by itself, the first thing I wanna do is subtract pie from both sides. Now we'll cancel over here leaving you with two, theta is equal to two pi N minus pi I can't really simplify that much. So I can go ahead and fully isolate the by dividing both sides by two. Now that will cancel over here leaving me with theta is equal to two pi N minus pi divided by two. Now, we can simplify this a little bit further. This is technically correct because the is by itself, but we can further simplify this because two pi N divided by two will just leave me with pi times N and then I'm subtracting pi over two. So my final answer is that theta is equal to pi N minus pi over two. And that represents all of my solutions to my original equation. Now, even though this can get a little bit complicated, you're always just going to follow your steps and you will get to the right answer. Thanks for watching and let me know if you have any questions.