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

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### 1 Answer

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, .....**