INSTRUCTOR: Consider this equation-- sine x is equal to 1/2. What is the value of x? So sine of what angle is equal to 1/2? If you use a unit circle, you know that sine of 30 degrees, or pi/6, is equal to 1/2. And sine is positive in quadrants I and II. 

So sine 5 pi over 6 is also equal to 1/2. So therefore, you can say x is equal to pi/6, and x is equal to 5 pi over 6. But if you want to find all solutions, it doesn't end there. 

For example, if you add 2 pi to pi/6, that would give you 13 pi over 6. Sine of 13 pi over 6 is also 1/2. And if you add 2 pi to 5 pi/6, that will give you another coterminal angle. 2 pi is the same as 12 pi over 6. So 5 pi over 6 plus 12 pi over 6 is 17 pi over 6. 

So to write an equation that gives us all solutions, it's going to be pi/6 plus 2 pi n. The period of sine and cosine is 2 pi, where n is any integer. So n could be 0 plus or minus 1, plus or minus 2, and so forth. And so these are the two answers. So any time you need to find all solutions, make sure you write your answer in terms of n. 

Let's work on another example. So let's say that cosine x is equal to negative square root 2 divided by 2. Find all answers that satisfies this equation. So cosine of what angle is equal to negative square root 2/2? Now, keep in mind, cosine is negative in quadrants II and III. 

And you need to know the reference angle. Cosine 45 is equal to the square root of 2/2. So in quadrant II, an angle of 135 produces a reference angle of 45. And in quadrant III, 225 produces a reference angle of 4-- of 45. So therefore, cosine of 135 degrees, or cosine 3 pi over 4, that's equal to negative square root 2/2. 

And also, cosine 5 pi over 4 is equal to the same value. So we can write the solution as this-- x is equal to 3 pi divided by 4 plus 2 pi n, where n is any integer. And x is also equal to 5 pi over 4 plus 2 pi n. 

Here's another one. Let's say we have this equation 2 sine x plus the square root of 3. Let's say it's equal to 0. And this time, we're going to restrict our answers from 0 to 2 pi. So when you see this, that means you don't need to add 2 pi n to your answer. You only want the answers in this range. 

So first, let's isolate sine x. And we could do so by subtracting both sides by the square root of 3. So 2 sine x is equal to negative square root 3. And now let's divide both sides by 2. So sine x is equal to negative square root 3 over 2. 

Now first, let's find the reference angle. If we draw the 30, 60, 90 triangle, across 30 is 1. Across 60 is root 3. And across the hypotenuse is 2. Now, the angle has to be 60 because sine is equal to opposite, which is root 3, divided by the hypotenuse. 

So we know that sine 60, or sine pi/3, is equal to root 3/2. But now we need the angle in quadrants III and IV. That's where sine is negative. 240, or 4 pi over 3, has a reference angle of 60. So sine of 4 pi over 3 is equal to this value, and also 5 pi over 3 have the same reference angle as pi/3. 

And so that will give you that answer. So x is equal to the values inside x as an angle. So x is 4 pi over 3 and 5 pi over 3, which is in this range. So these are the answers to this problem. 

Here's another one that you could try-- 3 tangent x plus the square root of 3 is equal to 0. Find all solutions. So that means you want to write your answer with n. So what should we do? 

The first thing we need to do is subtract both sides by the square root of 3. So 3 tan x is equal to negative root 3. Next, let's divide both sides by 3. So now we have this expression-- tangent x is equal to negative square root 3 divided by 3. 

So what reference angle will give us this value? If we draw the 30, 60, 90 triangle, we need to use the 30-degree angle. Tangent of 30 is equal to the opposite side 1 divided by the adjacent side. 

Now, 1 over root 3 if you rationalize it, is equivalent to the square root of 3 divided by 3. And so we have a reference angle of 30. However, tangent is not negative in quadrant I. Tangent is negative in quadrant II and in quadrant IV. 

So 150 will give us the angle in quadrant II. And to get the angle in quadrant IV, that's 330. So tangent of 150, which is the same as 5 pi over 6, is equal to negative square root 3/3. And tangent, 11 pi over 6, which is 330, that's equal to the same thing. 

Now, we said that the period of sine and cosine is 2 pi. The period of tangent and cotangent is pi. So to write the answers, you can write it like this-- x is equal to 5 pi over 6 plus pi n as opposed to 2 pi n. And it's also equal to 11 pi over 6 plus pi n, where n is any integer. And so these are the answers. And make sure to make some sort of statement that defines n. 

Here's another problem. Let's say that 2 sine squared minus 1 is equal to 0. Find all solutions. So first, we need to add 1 to both sides. And so 2 sine squared is equal to 1. 

Next, we need to divide both sides by 2. So sine squared is equal to 1/2. So now, what should we do at this point? Our next step is to take the square root of both sides. 

The square root of sine squared is simply sine x. But now, what is the square root of 1/2? The square root of 1 is 1, and the square root of 2 is the square root of 2. And it's going to be plus or minus. Because, once you square something, it's always going to be positive. 

For example, looking at the expression sine squared is equal to 1/2. if I plug om positive 1 over square root 2 or negative 1 over square root over 2, it would still give me positive 1/2. Because when you square a positive number or a negative number, it will give you a positive result. 

But now let's simplify 1 over root 2. What we can do is rationalize it. So what we have now is plus or minus the square root of 2 divided by 2. So sine x can equal positive root 2/2 or negative. So therefore, instead of having two quadrants, the answer lies in all four quadrants. 

So sine is positive in quadrants I and II. So therefore, x could be 45 degrees or pi/4. And it could be 3 pi over 4. Now, sine is negative in quadrants III and IV. So it could be 5 pi over 4 and 7 pi over 4. 

Now, to write an expression to get all solutions, we need to add 2 pi n to every answer. So you can write your answer like this if you want to. Now you can simplify this since we have all four answers. 

Notice that each of these answers differ by pi/2. pi/4, which is 45, differs from negative 3 pi over 4 by 90 degrees. 135 minus 45 is 90. And if you add 90 to 135, you're going to get 225. If you add 90 to 225, you get 315. 

And so because each of these answers differ by 90 degrees, we can write one statement that represents all of these solutions. So x is pi/4 plus pi over 2n, and where n is any integer. So this will give us any answer. If n is 0, then we're just going to get pi/4. If n is 1, we'll get pi/4 plus pi/2, which is 3 pi over 4. If n is 2, we'll get pi/4 plus pi, which is 5 pi over 4. 

So this expression will give us all answers. But if you don't want to write it like this, you can just add 2 pi n to each answer. And mathematically, it will still be correct. You just have to do more writing.