|

Trignometry

New kid on the Block

 Joined: 13 Aug 2011 Post: 8
14 Aug 2011 12:33:28 IST
0 People liked this
4
371
solve the equation
Engineering Entrance , JEE Main , JEE Advanced , Mathematics , Trigonometry

ecos x=e-cos x + 4

Forum Expert
Joined: 19 Feb 2009
Posts: 1979
14 Aug 2011 14:07:21 IST
0 people liked this

Hello !
ecos x=e-cos x + 4
Let  ecos x   = t

Then  t = t^-1 + 4
t = 1/t  + 4
t^2 = 1 + 4t
t^2 -4t - 1 = 0

t = 2 + root(5)                       and t = 2 - root(5)

As t =
ecos x   and cosx has a maximum value of 1 ..so maximum value of t must be equal to t = e which is approximately 2.733 something ...and minimum value of t = 1/e  which is in fraction and positive ...
So t = 2+root(5) must not be the answer ..because it become greater than three ...(not allowed because tmax = 2.733)
And t = 2-root(5) is neglected because it become negative .....and we know tmin is positive ...!

So the given equation has zero solution :)

New kid on the Block

Joined: 13 Aug 2011
Posts: 8
15 Aug 2011 19:29:11 IST
0 people liked this

hi... but will u explain y u have used cap after t in 4th step.... nd plz. explain the logic behind it...

New kid on the Block

Joined: 13 Aug 2011
Posts: 8
15 Aug 2011 19:32:50 IST
0 people liked this

nd i also have problem regarding finding the solutions in trigonometry... hw to solve it?

Forum Expert
Joined: 19 Feb 2009
Posts: 1979
15 Aug 2011 22:11:12 IST
0 people liked this

*cap* is used for power notation ..!

t^2  means ...t square
t^-1 means ... t to the power -1 ..

Since i have assume  ecosx   as t   so   e-cosx will be equal to  (ecosx )-1 = t-1

And the substitute these values in equation again ...   t = t-1 + 4    and then the solution ..(I am not going to solve this again in the same step)

Logic is ...The assumption ...whenever u feel there can be a quadratic equation after assuming some constant ..go for it...I am highlighting few examples ...

1)  sinx^2 + 5cosx + 3 =  0

Here u can assume cosx = t  =>  1-cos^2x + 5cosx + 3 = 1-t^2 + 5t + 3 = 0
or  t^2 -5t - 4 = 0   ( and then further u can solve it )  and after getting the value of t suppose u get ..

t=0.5  and t = -3    ...so in this case t=-3 will be neglected because ...as we have assumed t equals to cosx ..which means t can not be less than -1 and greater 1 ...so definitely t=-3 ..doesn't fall in the range ..!

Hence the acceptable answer is t=0.5 ..

2) ln x +  2/lnx  = 4

Now how can u solve this without substituting ....Surely assume lnx = t

t + 2/t = 4   again a quadratic equation and ....blah blah blah..(the same step as i have posted above)

3) 2cosx   +  2-cosx   = 5

Again ....say  2cosx   = t  ...and further the same step ..!

I hope it is clear to u now ..

nd i also have problem regarding finding the solutions in trigonometry... hw to solve it?

The only way to master yourself in trigonometry is hard practice ...go for examples ...first ..solve as many as question as u can ..in begining u will find difficult to solve it..don't worry ...solve the same question again and again if u will something illogical in it....don't finish unless u understand the logic and steps completely ..

A.Das Gupta for trigonometry is i guess the best book covering and giving a handful knowledge about the topic...Further when u feel confidence ..Go for standard book like M.Khanna ...and Arihant ..and do some tougher problems too ..!

Good luck

 Some HTML allowed. Keep your comments above the belt or risk having them deleted. Signup for a avatar to have your pictures show up by your comment If Members see a thread that violates the Posting Rules, bring it to the attention of the Moderator Team

## For Quick Info

Name

Mobile

E-mail

City

Class

### Find Posts by Topics

Physics

Topics

Mathematics

Chemistry

Biology

Institutes

Parents Corner

Board

Fun Zone

Vertical Limit

Top Contributors
All Time This Month Last Week
1. Bipin Dubey
 Altitude - 16545 m Post - 7958
2. Himanshu
 Altitude - 10925 m Post - 3836
3. Hari Shankar
 Altitude - 9960 m Post - 2185
4. edison
 Altitude - 10815 m Post - 7797
5. Sagar Saxena
 Altitude - 8625 m Post - 8064