4sin(x)cos(x)+2sin(x)-2cos(x)-1=0
Having trouble solving this one. I was able to solve it partially by adding the 2cos(x)-1 to the other side, then factoring the rest. To get this:
2sin(x)(2cos(x)+1)=2cos(x)+1
then divide both sides by 2cos(x)+1, we get 2sin(x)=1 solve for it solutions between 0

Question

4sin(x)cos(x)+2sin(x)-2cos(x)-1=0
Having trouble solving this one. I was able to solve it partially by adding the 2cos(x)-1 to the other side, then factoring the rest. To get this:
2sin(x)(2cos(x)+1)=2cos(x)+1
then divide both sides by 2cos(x)+1, we get 2sin(x)=1 solve for it solutions between 0<x<2pi and x = pi/6, 5pi/6. However in the back of the book the answers are pi/6, 5pi/6, 2pi/3, 4pi/3.

how do i solve to get all the answers? thank you.

asnaseer · Answer

instead of dividing both sides by 2cos(x)+1, move that expression from the right-hand-side to the left-hand-side and factorise

asnaseer · Answer

$$2\sin(x)(2\cos(x)+1)=2\cos(x)+1$$$$\therefore 2\sin(x)(2\cos(x)+1)-(2\cos(x)+1)=0$$

asnaseer · Answer

factorise that expression

asnaseer · Answer

by dividing both sides by 2cos(x)+1 you were effectively throwing away a set of solutions

mathmale · Answer

@asnaseer :  good guidance on your part.  Please let @physicscrap try the actual factoring and then respond.  Thanks.

asnaseer · Answer

that was my plan @mathmale :)

asnaseer · Answer

otherwise the asker will never learn :)

anonymous · Answer

wow that was fast, thanks guys. got it.

asnaseer · Answer

yw :)