October 10th, 2008, 05:33 PM  #1 
Newbie Joined: Oct 2008 Posts: 8 Thanks: 0  PreCalc 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 
October 10th, 2008, 08:14 PM  #2 
Newbie Joined: Oct 2008 Posts: 4 Thanks: 0  Re: PreCalc 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).

October 11th, 2008, 07:52 AM  #3 
Newbie Joined: Oct 2008 Posts: 8 Thanks: 0  Re: PreCalc Trig equation
As stated by the title and the fact that you have to solve for x, these are clearly Trig equations.

October 11th, 2008, 08:41 AM  #4 
Senior Member Joined: Jul 2008 Posts: 895 Thanks: 0  Re: PreCalc 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. 
October 11th, 2008, 03:15 PM  #5 
Global Moderator Joined: Dec 2006 Posts: 20,633 Thanks: 2080 
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. 
October 12th, 2008, 04:10 PM  #6 
Newbie Joined: Oct 2008 Posts: 8 Thanks: 0  Re: PreCalc Trig equation
for 4d, I don't get how your 2sin^2(x) becomes 1cos(x) on the bottom of the right side.

October 14th, 2008, 01:26 AM  #7 
Global Moderator Joined: Dec 2006 Posts: 20,633 Thanks: 2080 
Not 1  cos(x)! It became 1  cos(2x) due to a wellknown identity.

October 14th, 2008, 04:17 PM  #8 
Newbie Joined: Oct 2008 Posts: 8 Thanks: 0  Re: PreCalc Trig equation
Cool, I learned something. Never thought to use it in that way.


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