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anonymous
 4 years ago
Solve by factoring: 2sinxcosx=sinx in [0,2π)
anonymous
 4 years ago
Solve by factoring: 2sinxcosx=sinx in [0,2π)

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ganeshie8
 4 years ago
Best ResponseYou've already chosen the best response.1begin by subtractcting sinx both sides, and then factor sinx

ganeshie8
 4 years ago
Best ResponseYou've already chosen the best response.1\(2\sin x \cos x=\sin x \) subtract sinx both sides \(2 \sin x \cos x  \sin x = 0\) factor out sinx \(\sin x (2 \cos x  1) = 0 \)

ganeshie8
 4 years ago
Best ResponseYou've already chosen the best response.1We now have two factors whose product is zero, so the original equation will be satisfied when either factor is zero.

ganeshie8
 4 years ago
Best ResponseYou've already chosen the best response.1first set first factor = 0 => \(\sin x = 0\) the sin function 0, when \(x = 0 \) or \(x = \pi\) or \(x = 2\pi\)

ganeshie8
 4 years ago
Best ResponseYou've already chosen the best response.1similarly set the second factor = 0, and try getting other remaining solutions.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I think i get this.. so the final answer would be x=0 or x=pi

ganeshie8
 4 years ago
Best ResponseYou've already chosen the best response.1not exactly thats half of the solutions only.. you need to set the second factor also equal to 0, and see what solutions u get

ganeshie8
 4 years ago
Best ResponseYou've already chosen the best response.1set 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 : \(x = 2\pi  \pi/3 = 5\pi/3\)

ganeshie8
 4 years ago
Best ResponseYou've already chosen the best response.1so, the solutions in interval \([0, 2\pi]\) are : \(0, \ \pi, \ \pi/3, \ 5\pi/3, \ 2\pi\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Got ya! I just got all of those except for 2pi..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0But yes makes perfect sense!

ganeshie8
 4 years ago
Best ResponseYou've already chosen the best response.1dw:1352555743690:dw

ganeshie8
 4 years ago
Best ResponseYou've already chosen the best response.1since 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.
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