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October 10th, 2008, 05:33 PM   #1
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Pre-Calc Trig equation

I can't solve this questions nor can my teacher. We are just starting trig equations, so I'm quite new to this as well. Help to the starting steps to solve these would be greatly appreciated.

EDIT OCT11: I got em all, but I'd just like to see other ways of solving that don't involve dividing.

4. Solve for x in the given interval
(d)

tan (2x) = 8cos^2(x) - cot (x) , [0,pi/2]

(e)

tan(x) + sec (2x) = 1 , [-pi/2, pi/2]

6. If tan(2x) = -24/7 , find the values of sin x and cos x
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October 10th, 2008, 08:14 PM   #2
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Re: Pre-Calc Trig equation

Are these trigonometric identities? If so, I can solve these (I'm a tad baffled as to how your professor doesn't know).
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October 11th, 2008, 07:52 AM   #3
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Re: Pre-Calc Trig equation

As stated by the title and the fact that you have to solve for x, these are clearly Trig equations.
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October 11th, 2008, 08:41 AM   #4
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Re: Pre-Calc Trig equation

Unless I see some personal effort, it's just another "do my homework for me" question. All it takes is to show some work ...what you are expecting of those willing to help. At the very least you could expand the functions of 2x to functions in x. Then you can in fact be helped. How else do we tell the difference between an honest query, a "do my homework", or a troll? If you "got em all", it is simple to upload your effort here. This site allows attached files.

If your teacher can't do these, make every effort to get him/her replaced. It' your money and taxes paying for this education.

P.S. To distinguish between the two: an equation is true for some values of a variable [or none]. An identity is true for all values.
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October 11th, 2008, 03:15 PM   #5
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4. (d) $\displaystyle \tan (2x) = 8\cos^2(x) - \cot (x) = 8\cos^2(x) - \frac{2\sin(x)\cos(x)}{2\sin^2(x)}$
Hence $\displaystyle \frac{\sin(2x)}{\cos(2x)}=8\cos^2(x)-\frac{\sin(2x)}{1-\cos(2x)},$
and so $\displaystyle \frac{\sin(2x)}{\cos(2x)}-\sin(2x)=8\cos^2(x)(1-\cos(2x))-\sin(2x),$
i.e., $\displaystyle \frac{\sin(2x)}{\cos(2x)}=4(1+\cos(2x))(1-\cos(2x))=4(1-\cos^2(2x))=4\sin^2(2x).$
It's fairly easy to finish off from there. There are three solutions in the given range.

(e) $\displaystyle \tan(x)+\sec(2x)=1\;\Rightarrow\; 2\sin(x)\cos(x)=2\cos^2(x)(1-\sec(2x)),$
i.e., $\displaystyle \sin(2x)=(\cos(2x)+1)(1-\sec(2x)),$
so $\displaystyle \tan(2x)=(1+\sec(2x))(1-\sec(2x))=1-\sec^2(2x)=-\tan^2(2x).$
It's fairly easy to finish off from there. There are three solutions in the given range.

Last edited by skipjack; October 20th, 2015 at 10:32 AM.
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October 12th, 2008, 04:10 PM   #6
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Re: Pre-Calc Trig equation

for 4d, I don't get how your 2sin^2(x) becomes 1-cos(x) on the bottom of the right side.
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October 14th, 2008, 01:26 AM   #7
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Not 1 - cos(x)! It became 1 - cos(2x) due to a well-known identity.
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October 14th, 2008, 04:17 PM   #8
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Re: Pre-Calc Trig equation

Cool, I learned something. Never thought to use it in that way.
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