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## Precalculus

### Course: Precalculus>Unit 2

Lesson 7: Sinusoidal equations

# Solving sinusoidal equations of the form sin(x)=d

Sal finds the expressions that together represent all possible solutions to the equation sin(x)=1/3. Created by Sal Khan.

## Want to join the conversation?

• At , is there some reason that -0.34+pi*n is left out of the answer? It seems like it belongs as well. •   Well at Sal explains why -0.34+2pi*n won't work: sin(-0.34)= -1/3 and not 1/3 and if we add a multiple of 2pi we are gonna wind up there again. But if we add n pi to it there are two cases which might happen: we choose an uneven n and we are gonna end up with (pi - 0.34) + (n-1)pi, take sine of that: sin(2.80) = 1/3, that (n-1) is always gonna be even so it won't change anything, however when we chose an even n, we will get -0.34 + 2pi*n, now sine(-0.34) = -1/3, and we don't want that. So for half of the Ns we would get a wrong answer.
• Why am i getting 19.4712 when im taking sin^-1 (1/3) and not .3398 like sal https://www.youtube.com/watch?v=NC7iWEQ9Kug#t=172 ? (got a casio fx-9750 if that matter) •   19.4712 is the angle in degrees.
0.3398 is the angle in radians.
• Is it just me, or are these two videos not that useful for solving the problems given in the practice. I've missed every one of them so far, and feel like the videos failed to cover something. The explanations given when I click "tips" or whatever are mystifying. They mention identities that are not covered in these videos. • I don't understand this, previously Sal said that arcsin outputs can only be in the first and fourth quadrant, how come we take 2.8 radians as an answer if it's in the second quadrant? • Since x is the unknown value, we are looking for x. With inverse trig functions, x represents the domain of a "vanilla" sin function and the range of the inverse function: they are the same thing.
Now although we need to restrict the range of arcsin to get a valid inverse function, of course other angle options are true. We need to be careful to see what the question is asking us to find. The solution set of an equation will contain all possible answers for that equation.
https://youtu.be/NC7iWEQ9Kug?t=19 on the unit circle, the principal value is in QI, being in the range 'negate' 'pi/2' 'less than or equal to 'theta' 'less than or equal to' 'pi/2'.
• Do you tell the 1/3 is in radians just because there is no degree notation? • Yes. Basically, once you get introduced to radians, that is the end of using degrees. All higher math is done in radians only.

There is no unit in radians because it is just a ratio. The angle in radians is the ratio of the arc length to the radius.
• I'm confused as to when you're supposed to limit the range of the inverse trig functions. In this video, shouldn't 2.80 + 2pi(n) be invalid because sin^-1 is limited to [-pi/2, pi/2]? • The range of the arcsine function has to be bounded because otherwise the input would map to several outputs.
But this doesn't mean that the domain of the sine function must be bounded because several inputs can map to the same output.

So, if we are asked what arcsin(1) is, then the answer is 𝜋∕2,
but if we are asked what value of 𝑥 gives us sin(𝑥) = 1, then the answer would be 𝜋∕2 + 𝑛∙2𝜋, where 𝑛 is an integer.
• How would would solve something that is this complicating?
2sin^2x+cosx-1=0
(1 vote) • OK, hmmm
2sin²x + cos x - 1 = 0
The key here would be to use the trig identities.
The most basic one is cos²x + sin²x = 1
(This is sometimes called the Pythagorean Identity)
How does it help? It gets rid of the messy 2sin²x that makes your equation such a complicated thing to solve.
First solve the identity for sin²x
cos²x + sin²x = 1
sin²x = 1 - cos²x
so, 2 ( sin²x ) = 2(1 - cos²x)
Now we plug that into your challenge problem:
2(1 - cos²x) + cos x - 1 = 0
2 - 2cos²x + cosx - 1 = 0
- 2cos²x + cosx - 1 + 2 = 0
- 2cos²x + cosx + 1 = 0 → switch signs (multiply by -1)
2cos²x - cosx - 1 = 0
this is the same pattern as 2a² - a - 1, which factors into (2a + 1)(a - 1)
2cos²x - cosx - 1 = 0 factors into (2cosx + 1)(cosx - 1) = 0
Now you can solve using the zero property
(2cosx + 1) = 0; so 2cosx = -1 and cosx = -1/2
(cosx - 1) = 0; so cosx = 1
Now, finally, you can use arccos to solve for possible x values (measure of the angle x).
If cosx = -1/2, arccos (-1/2) = 120 degrees or 2π/3
If cosx = 1, arccos (1) = 0 degrees or 0 radians
The rest of the answer depends on exactly what they were asking--the principal value or all possible values.
Hope that helps
• hey we just set a limit to values of sin from pi/2 to 3pi/2 then why are we considering the angle in 2 quadrant at • sir, -0.34+(pi)n should also be an answer • I'm incredibly confused when it cones to solving these equations. I've been doing practice questions for over an hour trying to pick it apart.

First it asks me to select one or more expressions that together represent all solutions to the equation. And then it gives me sin(x)=0.4 to work with and a bunch of answers (e.g. 21.80​∘​​ +n⋅360​∘, 66.42​∘+n⋅180​∘) How do I know when to append the n*180 or n*360? When I click a hint, it shoots me a rang of -180∘ < x < 180.
Where did it get that range from? That's not even considering when I'm using radians. How can I tell if a decimal is in the range of, say, 3pi/2 and 9pi/2?
I'm incredibly confused where they are getting the ranges from, and I've never encountered these "identities" of sin, cos, tan, etc.
Any help would be greatly appreciated.
​​ • This might help.

Explanation:

So n is used to represent multiple solutions.

I presume when you are referring to 2pi radians you are referring to period.

Suppose we have the question: 3cos(2pi*t/3) + 10 = 11

Don't be overwhelmed if you don't understand where these theorems come from. They just come from the unit circle. Nothing is stopping you from applying these theorems even if you don't know the proof.

To solve these kinds of question we need to know the following:

The general solution for:

cos(theta) = a

is

2*n*pi +/- alpha (1)

__________________
sin theta = a

is

theta = n*pi +(-1)^n * alpha (2)

___________________
tan(theta) = a

is

theta = n*pi + alpha (3)

Of course you are not going to be able to apply theorem (1)/(2)/(3) in these types of questions because of the coefficent of t is 2pi/3 not 1.

So let u = 2pi/3 * t. With this substitution you can now apply the theorem as the coefficent is now 1 which is coefficent of theta.

So you have

u = n*pi +(-1)^n * alpha # note alpha = sin^-1(a)

2pi/3 * t = n*pi +(-1)^n * alpha # make t the subject and your done.

Note: To save time in a exam you don't need to actually substitute u however if you feel uncomfortable you may do so.
(1 vote)