Wednesday, April 16, 2014

4sinxcosx-1=2(sinx-cosx) find x

4sinx*cosx -1 =
2(sinx-cosx)


We will expand the
brackets.


==> 4sinxcosx -1 = 2sinx
-2cosx


Now we will move all terms to the left
side.


==> 2cosx -2sinx +4sinxcosx -1 =
0


==> 2cosx +4sinxcosx - 2sinx -1 =
0


Now we will factor 2cosx and
-1.


==> 2cosx( 1+ 2sinx) -( 2sinx+1) =
0


Now we will factor
(2sinx+1)


==> (2sinx+1) *(2cosx -1) =
0


==> 2sinx = -1 ==> sinx = -1/2
==>


 x= pi+pi/6 = 7pi/6 +
2npi


x = 2pi - pi/6 = 11pi/6 +
2npi


==> 2cosx -1 = 0 ==> cosx =
1/2


==> x = pi/3 +
2npi


==> x = 2pi - pi/3 = 5pi/3 +
2npi


==> x = { 7pi/6 + 2npi , 11pi/6 +
2npi, pi/3 + 2npi , 5pi/3+ 2npi} n= 1, 2, 3, 4,
.....

No comments:

Post a Comment