Another maths question! (trigonometric identities)

Bugz

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Fuck this unit is hard.

Anywho - any help with this?

Find to the nearest integer, the values of x in the interval 0 < x < 180 degrees for:

3 sin (squared) x - 7 cos 3x - 5 = 0


I've tried putting it into the quad. formula to find the 2 possible solutions and finding separate Primary angles for each but my answers don't match those in the back:

36, 84, 156 (all degrees).

ASTC applies here btw :)
 

Lamp

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LOL isn't it nice to know that once you've finished school most of us never have to bother with that shit for the rest of our lives
 

leviathane

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3 sin (squared) x - 7 cos 3x - 5 = 0


I've tried putting it into the quad. formula to find the 2 possible solutions and finding separate Primary angles for each but my answers don't match those in the back:
have you tried putting it into the same form? ie in cos form?
 

Jugvayne

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Are you sure that is the right equation? Try putting the values of x in; they don't work.
 

ileks

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We did this ages ago, and most of the time you would just do:

sin(squared)X = cos(squared)X - 1 or something

and substitute values in, factorise to get the different values of X. (also i think because its 7cos3x you would do 0<3x<540). Then use a CAST diagram
 

Jugvayne

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We did this ages ago, and most of the time you would just do:

sin(squared)X = cos(squared)X - 1 or something

and substitute values in, factorise to get the different values of X. (also i think because its 7cos3x you would do 0<3x<540). Then use a CAST diagram

The identity is sin^2 x = 1 - cos^2 x. You cannot however use that equation in this situation then factorise, cos x and cos 3x are completely different functions of x. It is possible to get a cubic in cos x since cos 3x = 4cos^3 x -3cos x, but this is pretty irrelevant since Bugz must have given the wrong equation or the wrong solutions from the book as they don't work in the equation. :p
 

ileks

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yeah when i tried i couldnt factorise because of the 7cos3x part, would have been ok if it was just 7cosx. /shrug
 

Bugz

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3 sin (squared) 3x - 7 cos 3x - 5 = 0

:d

Managed it ^^

My typing error owned me.
 

Jugvayne

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Ahh righteo. Well I'm guessing you used sin^2 3x = 1 - cos^2 3x

re-arranging that you should get a quadratic in cos 3x that should factorise nicely.

(3cos3x + 1)(cos 3x + 2)

Remember that one of the solutions will be invalid since |cos 3x| < 1.

Also remember when your inverse cos your valid root, you'll get two answers for 3x, one positive and one negative.

Say if you still can't get the given answers.

Edit: nevermind then :p I'm guessing you're on Core 2 or something, wait till the trig gets even more fun at Core 3 and 4, and even FP2
 

Bugz

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Ahh righteo. Well I'm guessing you used sin^2 3x = 1 - cos^2 3x

re-arranging that you should get a quadratic in cos 3x that should factorise nicely.

(3cos3x + 1)(cos 3x + 2)

Remember that one of the solutions will be invalid since |cos 3x| < 1.

Also remember when your inverse cos your valid root, you'll get two answers for 3x, one positive and one negative.

Say if you still can't get the given answers.

Edit: nevermind then :p I'm guessing you're on Core 2 or something, wait till the trig gets even more fun at Core 3 and 4, and even FP2

I didn't actually notice that factorisation there - so I used the quadratic formula.

And yes, currently C2. We're doing C1, C2 and M1 in January and C3, C4, M2 in June/July.
 

Golena

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I spent most of my A-level maths course in McDonalds instead..

I'm now starting to remember why!
 

Nate

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Sorry I'm only a qualified FH-Doctor, if you have any medical problems I can help you..but Math's? Fuck off!
 

Huntingtons

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Sorry I'm only a qualified FH-Doctor, if you have any medical problems I can help you..but Math's? Fuck off!

in a qualified FH-Paedo, if you have any children problems send the kids to me.
 

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