Solve for exact solutions over the interval [0°, 360°).
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Question

Solve for exact solutions over the interval [0°, 360°).

sin θ/2 = -√3/2

Accepted Answer

Welcome back. Everyone in this problem, we want to determine the exact solution of the trigonometric equation of the sine of 1/5 of theta equals negative a half. And we're considering the interval between open bracket, zero degrees, 360 degrees closed parentheses for our answer choices. A says it's 210 degrees. B says 21 degrees and 57 degrees. C says 42 114 186 258 and 330 degrees. And D says it's a null set. In other words that there's no solution. Now, let's think a bit about our equation here or s of 1/5 of theta equals negative half. OK? Well, here in order to figure out what theta is, we'll need to figure out what this anger is. So let's let X be equal to or rather let's let Y be equal to 1/5 of the, OK. Then that means the sine of X or the sine of Y. But I keep saying X, the sine of Y is going to be equal to negative or half. Now let's ask ourselves what angle or the sign of what angle equals negative a half? Well, since we're looking for the exact solution, we can use the unit circuit to help us. Now, if you recall that on the unit circle in the unit circle, we basically have a circle. Let me draw that a bit better. OK. We have our X and Y axis and our circle goes from one to negative one on both axes. OK? And here we're trying to figure out from our unit circle, we are in negative half would be. Now recall that on the unit circle, the Xr and Y coordinates are in the form cosine theta, sine, theta. OK. So here that means the sine of theta or in this case, the sine of Y is going to be negative or half. So it's going to be negative a half on the Y axis. So the question is which value is that? Well, recall that if we were to go to our unit circle here, we'd realize that the sign of 210 degrees OK equals negative a half. In other words, the Y value is negative a half at 210 degrees. And the Y value is also negative a half. OK. My body is also a negative a half at 330 degrees. This one right here, it's as if it was coming across, OK. So at 330 degrees and at 210 degrees. So this tells us, then this tells us since that's true that the sign of 210 degrees is equal to the sign of 330 degrees, which is equal to negative half. OK. So that means then, OK, that means that the sign of 1/5 of theater is going to be equal, OK? It's gonna be equal to 210 degrees and is also going to be equal or rather not the sign of 1/5 of the rather 1/5 of the, my apologies. 1/5 of theta is going to be equal to 210 degrees and 1/5 of theta is also going to be equal to 330 degrees. What does that mean for us? Well, if 1/5 of theta equals 210 degrees, that means theta is going to be equal to 210 multiplied by uh five, which is 1050 degrees. And now remember that um the, the, the, the sine, the period of the sine function is 360 degrees. So it's gonna be 1050 degrees plus 360 degrees multiplied by N where N is any integer. But now let's think about it. Remember that in our solution set, it was supposed to be between zero and 360 degrees. But here, if we're starting at 1050 then the could not be within that solution set. Whereas if we were to add multiples of 360 therefore, there is no or this solution or this result is not in the solution set. OK. Now, if we go to 1/5 of theta equals 330 degrees, then this tells us that theta is going to be equal to 330 degrees multiplied by five, which equals 1650 degrees plus 360 degrees. N but again, this is greater than 360. Therefore, it's not in the solution set. Thus, for our equation, there is no solution for theta between zero degrees and 360 degrees. Where zero degrees is inclusive, we have an empty set. Therefore, the correct answer is D thanks a lot for watching everyone. I hope this video helped.

Solve for exact solutions over the interval [0°, 360°).
sin θ/2 = -√3/2

Key Concepts

Solving Basic Trigonometric Equations

Unit Circle and Reference Angles

Interval Considerations and Back-Substitution

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