Linear Trigonometric Equations - Online Tutor, Practice Problems & Exam Prep

To solve trigonometric equations like sin(θ) = 1/2, identify solutions on the unit circle, such as θ = π/6 and θ = 5π/6. For general solutions, add 2πn to each solution. For more complex equations, isolate the trigonometric function, then revert to basic equations. For example, from 4sin(θ) - 3 = 1, isolate sin(θ) = 1 to find θ = π/2. Understanding these steps enhances problem-solving skills in trigonometry.

Introduction to Trig Equations

Hey everyone. In working with trig functions, you've likely come across a statement such as this one, the sine of theta equals 1 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 sine is equal to 1 half, which leads us to the conclusion that trig equations have multiple solutions. 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 of the solutions to trig 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 sine is equal to 1 half. So coming to our unit circle over here starting in that first quadrant, I know that for my angle pi over 6, the sine is equal to 1 half. So that represents a solution to my trig equation here, θ=π6. Now continuing around my unit circle going into quadrant 2, I know that for 5pi over 6, the sine is also equal to 1 half. So this represents yet another solution to my trig equation. Now in quadrants 3 and 4, I know that my sine values are going to be negative. So there are no other solutions in here. So pi over 6 and 5 pi over 6 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 trig equation? Well, for my trig equation here, the sine of theta equals 1 half, I already know that pi over 6 and 5 pi over 6 represent 2 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 6, the sine of 13 pi over 6 is also equal to 1 half. So that represents yet another solution to this equation. And then if I were to continue on for another rotation and reach what would now be 17 pi over 6, the sine of 17pi over 6 is also equal to 1 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? Well, 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 2 pi n to each of them, where 2 pi represents a full rotation around my unit circle and n is just an integer that represents how many times you're going around it. Now you might also see this written as 2 pi k depending on your professor or your textbook, but it means the same exact thing. So let's look at our sineθ=1/2 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 2 solutions of pi over 6 and 5 pi over 6 that make this statement true. The sine of it is equal to 1 half. Now that we have these solutions that are on our unit circle, all that's left to do is add 2 pi n to each of them. And we've now covered all of our bases. This represents all of the possible solutions to this equation, the sine of theta 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 that 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 2 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.

Example 1

Hey everyone. In this problem, we're asked to find all solutions to the equation the tangent of theta is equal to the square root of 3. 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 2πn to each of those solutions. So, starting on my unit circle, I want to find all the values for which my tangent is root 3. So in quadrant 1, I know that for my angle π3, my tangent is equal to the square root of 3. Now also knowing that in quadrant 3, all of my tangent values are positive here and I'm looking for positive square root of 3, I know that from my angle 4π3, my tangent is also equal to the square root of 3. So here, two answers that I get are π3 and also 4π3. Now we can just go ahead and add 2πn to each of these solutions in order to get all of the solutions to this equation. So if we take both of these, add 2πn, this represents all of the solutions to our equation. But something that you actually might notice here is that π3 and 4π3 are exactly π radians away from each other. So, if I were to just take π3 and add π to it, I would still get all of my solutions. So starting with π3, if I add one rotation of π, if I add another one and then another one and then another one, I will still get all of my solutions. So while π3+2πn and 4π3+2πn still represent all of my solutions, if I want to further simplify this, I can write this as π3+πn. And this represents the most simplified form of all of my solutions because when working with the tangent, if I have two angles for which this is true, they're actually always going to be π radians 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 π radians away from each other, you can actually simplify this further. So our final answer here, in its most simplified form, is π3+πn for the tangent of theta being equal to the square root of 3. Thanks for watching, and I'll see you in the next one.

Find all solutions to the equation.

$\cos x=1$

• A. $x=0$
• B. $x=2\pi n$
• C. $x=\pi+2\pi n$
• D. $x=\frac{\pi}{2}+2\pi n,x=\frac{3\pi}{2}+2\pi n$

Find all solutions to the equation.

$\tan\theta=1$

• A. $\theta=2\pi n$
• B. $\theta=\frac{\pi}{4}+2\pi n$
• C. $\theta=\frac{\pi}{4}+\pi n$
• D. $\theta=\frac{5\pi}{4}+2\pi n$

Find all solutions to the equation.

$\sin\theta=-\frac{\sqrt3}{2}$

• A. $\theta=\frac{\pi}{3}+2\pi n,\frac{2\pi}{3}+2\pi n$
• B. $\theta=\frac{7\pi}{6}+2\pi n,\frac{11\pi}{6}+2\pi n$
• C. $\theta=\frac{4\pi}{3}+2\pi n,\frac{5\pi}{3}+2\pi n$
• D. $\theta=\frac{\pi}{6}+2\pi n,\frac{5\pi}{6}+2\pi n$

Example 2

Everyone. In this problem, we are asked to find all of the solutions to the equation the sine of theta equals 12. 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 sine is equal to 12. Now coming over here to my unit circle, I know that the sine of π6 is 12, and then also over here in quadrant 2, that's the other quadrant for which my sine values are positive. I know that the sine of 5π6 will also be positive 12. So my solutions on my unit circle are theta equals π6 and theta equals 5π6. Now I'm going to stop here and go ahead and convert these into degrees. Now these are radian measures. So whenever we have radian measures, we can simply multiply them by 180 degrees over π in order to convert them to degrees. So let's go ahead and do that to both of these radian angle measures. So I want to multiply this by 180°/π and then multiply 5π6 also by 180°/π. Now for that π6 times 180 over π, those πs are, of course, going to cancel, leaving me with 180 divided by 6. Now this gives me a value of 30 degrees. Now for 5π6, again, those πs are going to cancel, and I'm going to be left with 5 times 180 over 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 radian angle measure by 180 over π. Now from here, we have our angle measures in degrees. But remember, we want to find all of the solutions here. So we want to add multiple rotations to these. Now typically, we add 2πn, but 2πn is in radians. So instead of 2π, I want to express a full rotation in degrees. So instead of 2πn, 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 360n and 150 degrees plus 360n. Thanks for watching, and I'll see you in the next one.

How to Solve Linear Trigonometric Equations

Hey, everyone. We just learned how to find all of the solutions to a basic trigonometric 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: 4sin(θ)−3=1. 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 more complicated trigonometric equation and turn it right back into a basic trigonometric 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 trigonometric equations, and soon you'll be able to solve any linear trigonometric 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 trigonometric equation, we can do the exact same thing. But instead of isolating a variable, we're instead going to be isolating our trigonometric function as though it is a variable. So in our linear trigonometric equation that we have here, 4sin(θ)−3=1, we're going to treat our trigonometric function, the sine of theta, as though it is a variable. So this sine of theta is just like x. So since this is the same exact equation as the one we had over here, just now with a sine instead of an x, we can apply the exact same operations in order to isolate that trigonometric function ending up with sine of theta equals 1. Now we're back to a basic trigonometric equation that we can solve by using the unit circle, and we would end up getting that our angle theta is equal to π2. So all that we're doing here is isolating our trigonometric function and then applying what we already know.

Now, not all linear trigonometric equations will be this straightforward. So here, I'm going to walk you through some steps that will work for any linear trigonometric equation. Now looking at the example that we have here, we're given −2cos(θ)+3=0. And we're asked to find all of the solutions of this equation within 0 to 2π. So let's go ahead and get started with step 1 here, which is to isolate our trigonometric function using our basic algebraic operations. Now here our trigonometric 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 3 from both sides, canceling it on that left side, leaving you with −2cos(θ)=−3. Now to fully isolate this trigonometric function cosine of theta, I can divide both sides by negative 2, canceling that negative 2, leaving me with the cosine of theta is equal to positive root 3 over 2 with those negatives actually having multiple solutions. So in order to continue on here, we want to go ahead and continue on with step 2 and find all of the solutions that are on our unit circle. So here we have the cosine of theta is equal to root 3 over 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 π/6, the cosine is root 3 over 2. And continuing on around my unit circle, I also know that here in quadrant 4, at 11π/6, the cosine will be the square root of 3 over 2 as well. So my two solutions here on my unit circle are theta equals π/6 and theta equals 11π/6. So we have completed step 2. We have found all of the solutions on our unit circle, and we can continue on to step 3.

Now step 3 tells us that if our domain is not restricted, we want to go ahead and add 2πn 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 2π. Now, 0 to 2π 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 3 because my domain is actually restricted, so I don't have to worry about doing anything additional here.

Now for step 4, we're just left to isolate theta. Now, in our trigonometric function here, theta is already by itself in that argument of cosine, so I don't have to do anything for step 4 either. Theta is already isolated and by itself. So step 4 is done, and I have my final answers here: Theta equals π/6 and theta equals 11π/6. So now that we know how to solve linear trigonometric equations, let's continue to get some practice together. Thanks for watching, and I'll see you in the next one.

Find all solutions to the equation.

$3\sin\theta-6=-9$

• A. $\theta=\frac{\pi}{2}+2\pi n$
• B. $\theta=2\pi n$
• C. $\theta=\pi+2\pi n$
• D. $\theta=\frac{3\pi}{2}+2\pi n$

Find all solutions to the equation.

$\sqrt2\cos\theta+4=5$

• A. $\theta=\frac{\pi}{4}+2\pi n,\frac{3\pi}{4}+2\pi n$
• B. $\theta=\frac{\pi}{4}+2\pi n,\frac{7\pi}{4}+2\pi n$
• C. $\theta=\frac{3\pi}{4}+2\pi n,\frac{5\pi}{4}+2\pi n$
• D. $\theta=\frac{5\pi}{4}+2\pi n,\frac{7\pi}{4}+2\pi n$

Find all solutions to the equation.

$\sqrt3\tan\theta-7=-6$

• A. $\theta=\frac{\pi}{6}+2\pi n,\frac{5\pi}{6}+2\pi n$
• B. $\theta=\frac{5\pi}{6}+2\pi n,\frac{11\pi}{6}+2\pi n$
• C. $\theta=\frac{\pi}{6}+2\pi n,\frac{7\pi}{6}+2\pi n$
• D. $\theta=\frac{\pi}{6}+\pi n$

Example 3

Hey everyone! In this problem, we're asked to find all solutions to the equation 4 times the cosine of 2 theta plus pi plus 8 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 1, we want to go ahead and isolate our trigonometric function. In this case, our trigonometric function looks a little bit different. It is the full cosine of 2 theta plus pi. That's what we want to isolate. Now we can start by subtracting 8 from both sides, canceling over here, and leaving me with 4 times the cosine of 2 theta plus pi is equal to 12 minus 8, which is positive 4. Now from here, I can isolate this by dividing both sides by 4, canceling on that left side, leaving me just with my trigonometric function, cosine of 2 theta plus pi is equal to 4 divided by 4, which is positive 1. Now I've completed step number 1. My trigonometric function is isolated. It's all by itself. And I can move on to step number 2, finding all of my solutions on the unit circle.

This might seem a little bit strange to you at first because we're not just dealing with theta. We're dealing with 2 theta plus pi. But it really doesn't matter at this point in our process. It's not going to matter until later on in our steps because we're still trying to find a value for which the cosine is equal to 1. So, 2 theta plus pi, what does that need to be in order to get a cosine of 1? We're working through this the same way. So, coming to our unit circle here, where is my cosine going to be equal to 1? Well, I know that for 0, my cosine is 1, and that's actually the only angle for which that is true. So in order to get a cosine value of 1, that 2 theta plus pi needs to be equal to 0. So from here, we have completed step number 2. We found our solutions that are on our unit circle.

Now from here, if our domain is not restricted, we want to add \(2\pi n\) 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 \(2\pi n\) over here. So here I'm left with 0 + \(2\pi n\). Now, I can actually simplify that because that 0 isn't doing anything. So this is \(2\pi n\) on that right side. And then I'm left with 2 theta plus pi is equal to \(2\pi n\). That represents all of my solutions. But we have one final step here because we've completed step number 3, but step number 4 is to isolate theta. That's where we're actually going to deal with this 2 theta plus pi.

So from here, we're going to just go ahead and use algebra to isolate theta. In order to do that, I want this theta by itself. The first thing I want to do is subtract pi from both sides. Now it will cancel over here, leaving me with 2 theta is equal to \(2\pi n - \pi\). I can't really simplify that much. So I can go ahead and fully isolate theta by dividing both sides by 2. Now, that will cancel over here, leaving me with theta is equal to \(2\pi n - \pi\) divided by 2. Now we can simplify this a little bit further. This is technically correct because theta is by itself, but we can further simplify this because \(2\pi n\) divided by 2 will just leave me with \(\pi n\), and then I'm subtracting a \(\pi\) over 2. So my final answer is that theta is equal to \(\pi n - \frac{\pi}{2}\). 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.

Here’s what students ask on this topic:

How do you find all solutions to the trigonometric equation sin(θ) = 1/2?

To find all solutions to the equation sin(θ) = 1/2, start by identifying the angles on the unit circle where the sine value is 1/2. These angles are θ = π/6 and θ = 5π/6. For general solutions, add 2πn to each solution, where n is an integer. Thus, the complete set of solutions is θ = π/6 + 2πn and θ = 5π/6 + 2πn, where n ∈ ℤ.

How do you solve the trigonometric equation 4sin(θ) - 3 = 1?

To solve 4sin(θ) - 3 = 1, first isolate the trigonometric function. Add 3 to both sides to get 4sin(θ) = 4. Then, divide by 4 to obtain sin(θ) = 1. Next, find the angles on the unit circle where sin(θ) = 1, which is θ = π/2. For general solutions, add 2πn to this angle, where n is an integer. Thus, the complete set of solutions is θ = π/2 + 2πn, where n ∈ ℤ.

What are the steps to solve a linear trigonometric equation?

To solve a linear trigonometric equation, follow these steps: 1) Isolate the trigonometric function using algebraic operations. 2) Find all solutions on the unit circle for the isolated trigonometric function. 3) If the domain is not restricted, add 2πn (or πn for tangent and cotangent) to each solution, where n is an integer. 4) Ensure the variable is isolated. This method works for equations like 4sin(θ) - 3 = 1 or -2cos(θ) + √3 = 0.

How do you find all solutions to the equation cos(x) = -1?

To find all solutions to the equation cos(x) = -1, start by identifying the angles on the unit circle where the cosine value is -1. This angle is x = π. For general solutions, add 2πn to this angle, where n is an integer. Thus, the complete set of solutions is x = π + 2πn, where n ∈ ℤ.

How do you solve the trigonometric equation -2cos(θ) + √3 = 0?

To solve -2cos(θ) + √3 = 0, first isolate the trigonometric function. Subtract √3 from both sides to get -2cos(θ) = -√3. Then, divide by -2 to obtain cos(θ) = √3/2. Next, find the angles on the unit circle where cos(θ) = √3/2, which are θ = π/6 and θ = 11π/6. For general solutions, add 2πn to each angle, where n is an integer. Thus, the complete set of solutions is θ = π/6 + 2πn and θ = 11π/6 + 2πn, where n ∈ ℤ.