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

1. ganeshie8

begin by subtractcting sinx both sides, and then factor sinx

2. ganeshie8

$$2\sin x \cos x=\sin x$$ subtract sinx both sides $$2 \sin x \cos x - \sin x = 0$$ factor out sinx $$\sin x (2 \cos x - 1) = 0$$

3. ganeshie8

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

4. ganeshie8

first set first factor = 0 => $$\sin x = 0$$ the sin function 0, when $$x = 0$$ or $$x = \pi$$ or $$x = 2\pi$$

5. ganeshie8

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

6. johnnyalln

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

7. ganeshie8

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

8. ganeshie8

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 : $$x = 2\pi - \pi/3 = 5\pi/3$$

9. ganeshie8

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

10. johnnyalln

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

11. johnnyalln

But yes makes perfect sense!

12. johnnyalln

Thanks a lot!

13. ganeshie8

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14. ganeshie8

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.

15. ganeshie8

yw :)