Solve the equation sin 2Θ = 1, for 0 ≤ Θ

Question

Solve the equation sin 2Θ = 1, for 0 ≤ Θ < 2π .

Accepted Answer

Welcome back everyone. In this problem, we want to find the value of x where the function sine of 3x equals 1, where x is between 0 and 2 pi. Now for answer choices, a is the solution set a 6th of pi, 5 6th of pi, and 3 halves of pi. B is the solution set a half of pi and 3 halves of pi. C is the solution set a third of pi, 2 thirds of pi, and 3 halves of pi. And the d is the set a 6th of pi, 5 6th of pi, and 7 6th of pi. Now what do we know here? Well, we want to find the value of x for our solution where x is between 0 and 2 pi. We already know that on a unit circle, okay, recall that its coordinates are in the form the cosine of theta and the sine of theta where the sine of our angle is the y value and or the y coordinate and the cosine is the x coordinate. We also know that x is between 0 and 2 pi. But, what are we trying to find here? We are trying to find 3x in order to help us solve for x. So that tells us that 3x would have to be between 0 and 6 Pi. So 3x would be greater than or equal to 0 and less than 6 Pi. Now our circle has 2 Pi Revolutions. So in order for us to figure out all the possible values of x, we'll have to go through 3 revolutions that is 2pi, 4pi, and 6pi and figure out what all those values of 3x are when the sine of our angle 3x equals 1. So let's put that on our unit circle here, and we know that one on the y axis would be right there. So let's go through each revolution finding the value of 3x and then subsequently x. So let's start with the sine of 3x being equal to 1 where 3x is between 2 pi is less than 2 pi sorry but greater than or equal to 0. So, on that revolution if we notice here the sine of 3x being equal to 1 would mean that our angle would have to be a half of pi. Therefore, 3 x equals a half of pi. And if 3 x equals a half of pi, then that tells us that x equals a half of pi divided by 3, which is a 6th of pi. So that would be our first value of x for our first revolution. For our next revolution alright? So that would be the sine of 3x equals 1. Now we're looking at 3x being less than or equal to 4 pi, but greater than 2 pi. Then now we'd go from 2 pi to 3 pi and then back down to 4 pi. And at this point, halfway between 2 pi and 3 pi would be where our value of 3 x would be. And halfway between that would be 5 halves of pi. In other words, 2a half pi. So therefore, 3 x would be equal to 5 halves of pie. And if 3 x equals 5 halves of pie, then x would be equal to 5 half 5 halves of pie divided by 2 divided by 3, sorry, which would be equal to 5 6ths of pi. So, that's our second possible value for x. We're not done yet. We have one more revolution to make. So, I'm going to erase the 3pi and the 2pi and now, we're starting at 4pi. Put this one in red and then, we're going to go all the way around until we get to 6 pi. I think I could draw that a little bit better. Yes. That looks much better. And now we're going from 4 pi all the way to 6 pi. So now, looking at the sine of 3x equals 1 or the range between 4pi and 6pi, Then, that tells us that our angle would be this value right here. You notice the value is still the same but because it's the reference angle, we're looking at it for different revolutions. And now, in between that value, then our angle would be halfway between 4pi and 5pi. That is 3x would have been equal to 9 halves of pi. And if 3 x equals 9 halves of pi, then x would be equal to 9 halves of pi divided by 3, which is 9 6th of Pi. Or, we can simplify that because both of them have a common factor of 3. 3 goes into 9 3 times. 3 into 6, 2 times. So x would have been 3 halves of Pi. So, our possible answers so x equals 3 halves of Pi. Our possible answers would have been that either x equals a 6th of Pi, 5 6th of pi, or 3 halves of pi for x being between 0 and 2 pi. For these three values, indeed, x is in between 0 and 2 pi. So that gives us an idea that we're right. Therefore, our answer would have been answer choice a. Thanks a lot for watching everyone. I hope this video helped.

Solve the equation sin 2Θ = 1, for 0 ≤ Θ < 2π .

Key Concepts

Trigonometric Functions

Inverse Trigonometric Functions

Periodic Nature of Sine Function

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