## johnnyalln Group Title Solve by factoring: 2sinxcosx=sinx in [0,2π) one year ago one year ago

1. ganeshie8 Group Title

begin by subtractcting sinx both sides, and then factor sinx

2. ganeshie8 Group Title

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

3. ganeshie8 Group Title

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

4. ganeshie8 Group Title

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

5. ganeshie8 Group Title

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

6. johnnyalln Group Title

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

7. ganeshie8 Group Title

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 Group Title

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 Group Title

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

10. johnnyalln Group Title

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

11. johnnyalln Group Title

But yes makes perfect sense!

12. johnnyalln Group Title

Thanks a lot!

13. ganeshie8 Group Title

|dw:1352555743690:dw|

14. ganeshie8 Group Title

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 Group Title

yw :)