My Math Forum Pre-Calc Trig equation

 Trigonometry Trigonometry Math Forum

 October 10th, 2008, 05:33 PM #1 Newbie   Joined: Oct 2008 Posts: 8 Thanks: 0 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
 October 10th, 2008, 08:14 PM #2 Newbie   Joined: Oct 2008 Posts: 4 Thanks: 0 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).
 October 11th, 2008, 07:52 AM #3 Newbie   Joined: Oct 2008 Posts: 8 Thanks: 0 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.
 October 11th, 2008, 08:41 AM #4 Senior Member   Joined: Jul 2008 Posts: 895 Thanks: 0 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.
 October 11th, 2008, 03:15 PM #5 Global Moderator   Joined: Dec 2006 Posts: 20,966 Thanks: 2216 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: 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.
 October 14th, 2008, 01:26 AM #7 Global Moderator   Joined: Dec 2006 Posts: 20,966 Thanks: 2216 Not 1 - cos(x)! It became 1 - cos(2x) due to a well-known identity.
 October 14th, 2008, 04:17 PM #8 Newbie   Joined: Oct 2008 Posts: 8 Thanks: 0 Re: Pre-Calc Trig equation Cool, I learned something. Never thought to use it in that way.

 Tags equation, precalc, trig

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post Trigonometry 1 November 20th, 2012 12:45 PM mathkid Calculus 24 October 6th, 2012 11:19 PM trekster New Users 5 January 13th, 2012 02:13 PM miroslav_karpis Algebra 1 February 16th, 2009 07:00 AM squeeze101 Algebra 5 October 11th, 2008 05:58 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top