In Exercises 11–24, find all solutions of each equation.3 sin θ +...

Question

In Exercises 11–24, find all solutions of each equation.3 sin θ + 5 = ﹣2 sin θ

Accepted Answer

Hey, everyone in this problem, we're given the trigonometric equation. 13 sine C is equal to negative 12 sine C minus 25 for us to solve and to give the answer in ratings, we have four possible answer choices. OK? Option A is that C is equal to three pi divided by two plus pi N. Option B is that C is equal to three pi divided by two plus two pi N option C is that C is equal to pi divided by two plus pi N or three pi divided by two plus pi N. And option D is that C is equal to pi divided by two plus two pi N or three pi divided by two plus two pi N and in each of those cases N is an integer. So we're gonna start just by writing what we have. We have 13 sine C is equal to negative sine C minus 25. OK? And don't get overwhelmed and confused by having these sine functions in this equation. We're gonna treat it like any other equation. We wanna move everything with C to one side, all of the constant terms to the other side to try to isolate for C. So we can move our negative 12 C to the left by adding, we get 25 is equal to negative 25. We divide both sides by 25 get that sine C is equal to negative one. So there's a couple different ways you can think about this. You can remember the values of sine C, you could use a unit circle. That's another great option. What I'm gonna do is I'm gonna draw out a sign curve and we're gonna think about where we get this value of negative one. So if we draw a sign curve, remember that sign starts at 00, it increases to its maximum of one decreases back to zero at pi and then continues downwards to its minimum and then back up to Dubai. So the value of negative one we're gonna have is that this minimum value OK? And this is gonna occur at three pi divided by two. OK? Between pi and two pi. So we know that we have this value of three pi divided by two. We aren't given a specific interval. So we need to think about extending this graph. When does the next one occur? OK. And if we extend this graph, this dotted line can see that the next solution is gonna be two pi radiant away. OK. We have to go through a full cycle before we get to this minimum value, we only hit that minimum once every cycle. OK? And so the distance between these two solutions is gonna be two pi radiant. So if we want to write out our general solution or our full solution, OK? We can write that C is gonna be equal to three pi divided by two radiant plus two pi multiplied by N K or N is an in, OK? So we have that first value we found three pi divided by two. Then we have to add two pie to get to the next one. OK? Two pie to get to the next 12 pie to get to the next one. If we compare this to our answer choices, we can see that this corresponds with answer choice B that's it for this one. Thanks everyone for watching. See you in the next video.

In Exercises 11–24, find all solutions of each equation.3 sin θ + 5 = ﹣2 sin θ

Key Concepts

Trigonometric Functions

Algebraic Manipulation

General Solutions of Trigonometric Equations

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