Question

Solve by factoring: 2sinxcosx=sinx in [0,2π)

Answer 1

begin by subtractcting sinx both sides, and then factor sinx

Answer 2

\(2\sin x \cos x=\sin x \)

subtract sinx both sides

\(2 \sin x \cos x - \sin x = 0\)

factor out sinx

\(\sin x (2 \cos x - 1) = 0 \)

Answer 3

We now have two factors whose product is zero, so the original equation will be satisfied when either factor is zero.

Answer 4

first set first factor = 0

=>

\(\sin x = 0\)

the sin function 0, when \(x = 0 \) or \(x = \pi\) or \(x = 2\pi\)

Answer 5

similarly set the second factor = 0, and try getting other remaining solutions.

Answer 6

I think i get this.. so the final answer would be x=0 or x=pi

Answer 7

not exactly thats half of the solutions only.. you need to set the second factor also equal to 0, and see what solutions u get

Answer 8

set second factor = 0

=>

\(2 \cos x -1 = 0\)

\(\cos x = 1/2\)

since, cos is positive in first and fourth quadrants :
solution in first quadrant : \(x = \pi/3\)
solution in fourth quadrant : \(x = 2\pi - \pi/3 = 5\pi/3\)

Answer 9

so, the solutions in interval \([0, 2\pi]\) are : \(0, \ \pi, \ \pi/3, \ 5\pi/3, \ 2\pi\)

Answer 10

Got ya! I just got all of those except for 2pi..

Answer 11

But yes makes perfect sense!

Answer 12

Thanks a lot!