Thread: can't solve a trigo equ
View Single Post
  #2  
Old June 15th, 2008, 01:07 AM
red_dog's Avatar
red_dog red_dog is offline
MHF Contributor
  
Join Date: Jun 2007
Location: Medgidia, Romania
Posts: 1,203
Country:
Thanks: 22
Thanked 647 Times in 584 Posts
red_dog is a splendid one to beholdred_dog is a splendid one to beholdred_dog is a splendid one to beholdred_dog is a splendid one to beholdred_dog is a splendid one to beholdred_dog is a splendid one to beholdred_dog is a splendid one to behold
Default
Quote:
Originally Posted by afeasfaerw23231233 View Post
2sinx cos^2x - (1+sinx)(cos^2 x - sin ^2 x) = 0
I used the identities 2\sin x\cos x=\sin 2x and \cos^2x-\sin^2x=\cos 2x
Now, the equation becomes
\sin 2x\cos x-\cos 2x-\sin x\cos 2x=0\Leftrightarrow \sin x-\cos 2x=0
(I used the formula \sin(a-b)=\sin a\cos b-\sin b\cos a)
Using the formula \cos 2x=1-2\sin^2x the equation becomes 2\sin^2x+\sin x-1=0
Let \sin x=t and the equation 2t^2+t-1=0 has the roots t_1=-1, \ t_2=\frac{1}{2}
So you have to solve the equations \sin x=-1 and \sin x=\frac{1}{2}
Reply With Quote
The following users thank red_dog for this useful post:
Donate to MHF