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

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

Trigonometry
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begin by subtractcting sinx both sides, and then factor sinx
\(2\sin x \cos x=\sin x \) subtract sinx both sides \(2 \sin x \cos x - \sin x = 0\) factor out sinx \(\sin x (2 \cos x - 1) = 0 \)
We now have two factors whose product is zero, so the original equation will be satisfied when either factor is zero.

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first set first factor = 0 => \(\sin x = 0\) the sin function 0, when \(x = 0 \) or \(x = \pi\) or \(x = 2\pi\)
similarly set the second factor = 0, and try getting other remaining solutions.
I think i get this.. so the final answer would be x=0 or x=pi
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
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 : \(x = 2\pi - \pi/3 = 5\pi/3\)
so, the solutions in interval \([0, 2\pi]\) are : \(0, \ \pi, \ \pi/3, \ 5\pi/3, \ 2\pi\)
Got ya! I just got all of those except for 2pi..
But yes makes perfect sense!
Thanks a lot!
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since we are looking for \(\sin x = 0\), we look at the graph of \(\sin\), see that the graph of \(\sin\) is becoming \(0\) when \(x = 0\) or \(x = \pi\) or \(x = 2\pi\) since all these 3 values of x are in the range \([0, 2\pi]\) , all 3 values satisfy the equation, and interval.
yw :)

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