Math 1A/1B. Pre-Calculus: Solving Trigonometric Equations, Part 2

Let's solve trig equations. For example, let's find all solutions to this equation here. We'll begin by subtracting square root of 2 from both sides, which gives us 2 sine theta is equal to negative square root of 2. And now dividing by 2 gives us sine of theta is equal to negative square root of 2 over 2. Now remember with your unit circle, the sine of theta is the y-coordinate of the point of intersection of the terminal side of the angle and the unit circle. And looking down here in quadrant 3, we see that the y-coordinate of this point is negative root 2 over 2, which corresponds to the angle of 5 pi divided by 4. And also over here in quadrant 4, the y-coordinate is negative root 2 over 2, which corresponds to the angle of 7 pi divided by 4, which means both of these angles will have sine equal to negative root 2 over 2. So theta is equal to 5 pi divided by 4 and 7 pi divided by 4. But what else? Are these the only two angles that have sine equal to negative root 2 over 2? And they aren't, because if we go around this unit circle, the angle 2 pi radians in either direction will land on these same two points, which means that the sine of those angles will also equal negative root 2 over 2. And those angles are called coterminal angles because they share the same terminal side. For example, 5 pi divided by 4 plus 2 pi is a coterminal angle of 5 pi divided by 4. Or 5 pi divided by 4 plus 4 pi. Or 5 pi divided by 4 minus 2 pi. Or 5 pi divided by 4 minus 4 pi. And so on. All of these angles will have sine equal to negative root 2 over 2. The same thing with the coterminal angles of 7 pi over 4. Those angles will also have sine equal to negative root 2 over 2. So not only will these two angles have sine equal to negative root 2 over 2, but also any of their coterminal angles. And we write that in the following way. Theta is equal to 5 pi divided by 4 plus 2k pi where k is an element of z. This notation here means k is an integer. For example, when k is equal to 0, we get this angle of 5 pi divided by 4. And when k is equal to 1, we get this angle up here. When k is equal to 2, we get this angle. When k is equal to negative 1, we get this angle. When k is equal to negative 2, we get this angle. And so on. All of these angles will have sine equal to negative root 2 over 2. And we also have that 7 pi over 4 and any of its coterminal angles also have sine equal to negative root 2 over 2. So theta is also equal to 7 pi divided by 4 plus 2k pi where k is an integer. So any of these angles here will satisfy our equation. So these would be our answers. Alright, let's look at another example. Let's find all solutions to this equation. Well, we can begin by factoring out a cosine theta from the left-hand side which gives us cosine of theta times 2 cosine of theta plus 1 is equal to 0. So we have a product of factors equal to 0 which means either the first factor is 0, so cosine of theta is equal to 0 or the second factor is 0. So 2 cosine of theta plus 1 is equal to 0. Again, thinking of our unit circle and remembering that the cosine of theta is the x-coordinate of the point of intersection of the terminal side of the angle and the unit circle. And we see that up here the x-coordinate of this point is equal to 0 which corresponds to the angle of pi divided by 2. But also down here the x-coordinate is also 0 which corresponds to the angle of 3 pi divided by 2. And so for this first equation here the solution would be theta is equal to pi over 2, 3 pi over 2 or any of their coterminal angles. Which remember from the last example we can write that theta is equal to pi divided by 2 plus 2k pi where k is an integer. And theta is equal to 3 pi divided by 2 plus 2k pi where k is an integer. Now looking back over here at the unit circle in this type of situation where our two solutions are exactly pi units away from each other we don't necessarily have to write both of these statements over here. We can consolidate them into one statement realizing that we only need to add or subtract integer multiples of pi from pi halves to get to another solution to this equation. That is we can condense these two statements by writing theta is equal to pi over 2 plus k pi where k is an integer. Which are the solutions to the first equation. Now if you would have left both of these statements over here you would have been correct. This is just a more condensed way of writing the answer. Alright let's look at the solutions to this second equation here. Well we can subtract 1 and divide by 2 which gives us cosine of theta is equal to negative 1 half. Again remembering that the cosine of theta is the x coordinate of the point of intersection of the terminal side of the angle in the unit circle. So looking over here at our unit circle we see up here in quadrant 2 that the x coordinate of this point is negative 1 half which corresponds to this angle of 2 pi divided by 3. But also down here in quadrant 3 the x coordinate is also negative 1 half which corresponds to this angle of 4 pi divided by 3. Which means both of those angles have cosine equal to negative 1 half. But so do all of their coterminal angles. Which means the solutions to this second equation are given by theta is equal to 2 pi divided by 3 plus 2 k pi where k is an integer. And theta is equal to 4 pi divided by 3 plus 2 k pi where k is an integer. And since these solutions are not pi units away from each other we cannot consolidate them. And therefore the solution to our original equation are all of these angles here as well as all of these angles here. And this is how we solve trig equations. Thank you and we'll see you next time.