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# Trig. function problem Watch

1. Hi everyone,

I have been asked to solve the equation in the attachment, between 0 and 2 pi. I have an answer of 5/12 pi (which is right), but I thought 19/12 pi would solve it, but it doesn't.

Why is this - because the cos graph for radians says that it should?
Attached Images

2. (Original post by Electrogeek)
Hi everyone,

I have been asked to solve the equation in the attachment, between 0 and 2 pi. I have an answer of 5/12 pi (which is right), but I thought 19/12 pi would solve it, but it doesn't.

Why is this - because the cos graph for radians says that it should?
We have so this means that we have since, sqrt(3)/2 is positive:

.
3. Cool. So am I right in saying 5/12 pi is the only answer between 0 and 2 pi?
4. (Original post by Electrogeek)
Cool. So am I right in saying 5/12 pi is the only answer between 0 and 2 pi?
Well... not quite, remember that if is a solution, then so is . This gives x outside 2pi and 0, so we ignore it.

Buuut, is also a solution...
5. (Original post by Electrogeek)
Cool. So am I right in saying 5/12 pi is the only answer between 0 and 2 pi?
Whilst I've got a few minutes, I'll write up a thing.

So you know that solutions to trig equations are given by using the CAST method, we take the arctrig function of the absolute value of the RHS, in this case: .

Now our solutions are given by the infinite list, we can split this into two infinite lists, one is:

And the other is:

Side note: Getting the idea? The reason why this is so is because is a -periodic function so we have: this is true for as well - tangent is slightly different in the sense that it is a -periodic function, so and instead of adding to each solution, you'd add .

Anyways, the first few solutions are

So adding to both sides and throwing anything that isn't in our range gives us:
6. (Original post by Zacken)
Well... not quite, remember that if is a solution, then so is . This gives x outside 2pi and 0, so we ignore it.

Buuut, is also a solution...
Cool - I understand now. Thanks.
7. (Original post by Electrogeek)
Cool - I understand now. Thanks.
I typed up something above if you want to have a look.
8. (Original post by Zacken)
I typed up something above if you want to have a look.
So... would these be right for these equations? (answers are in the attachment)
Attached Images

9. (Original post by Electrogeek)
So... would these be right for these equations? (answers are in the attachment)
Assuming you mean between 0 and 2pi, I agree with the one for but not for you should have four solutions for the sin one.
10. (Original post by Zacken)
Assuming you mean between 0 and 2pi, I agree with the one for but not for you should have four solutions for the sin one.
Are the two given for sin right so far?

Or is is -pi/12 (+/- pi), and 7/6 pi (+/- pi)?
11. (Original post by Electrogeek)
Are the two given for sin right so far?

Or is is -pi/12 (+/- pi), and 7/6 pi (+/- pi)?
Should be

and then throw away everything you don't want.
12. (Original post by Zacken)
Should be

and then throw away everything you don't want.
Thank you! Makes a lot more sense now.
13. (Original post by Electrogeek)
Thank you! Makes a lot more sense now.
No problem.

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