## lgbasallote 3 years ago Find all values of x such that $$\sin 2x = \sin x$$ and 0 < x < 2$$\pi$$

1. lgbasallote

i suppose the only way for this to happen is if the x were 0....

2. JamesWolf

yeah or pi, but you specified x < 2pi

3. lgbasallote

hmm i suppose pi works

4. lgbasallote

sin 2x = sin x arcsin both sides 2x = x 2x - x = 0 x = 0 how can i get the other values?

5. JamesWolf

hmmm good question

6. leedomathgeek

sin(2x)=2sin(x)cos(x) Therefore, sin(2x)=sin(x) --> 2sin(x)cos(x)=sin(x) 2sin(x)cos(x)-sin(x)=0 sin(x)[2cos(x)-1]=0 Now either sin(x)=0 ---> x=0,pi,... or 2cos(x)-1=0 --> cos(x)=1/2 ---> x=pi/3

7. lgbasallote

then i suppose the others can be solved by adding pi?

8. leedomathgeek

Why do you want to add a pi?

9. lgbasallote

because one angle is missing

10. leedomathgeek

You have only 3 solutions

11. leedomathgeek

0, pi/3, pi

12. lgbasallote

there's actually 5.. but i know how to get one of the missing angles...

13. leedomathgeek

oh ya i though x between 0 and pi

14. leedomathgeek

sure 2pi is the 4th solution

15. lgbasallote

yes

16. leedomathgeek

and x=2pi-p/3=5pi/3

17. leedomathgeek

is the 5th solution

18. lgbasallote

oh subtract pi/3 from 2pi...ah yes... the negative angle

19. leedomathgeek

exactly ...since cos(x) is positive in the 4th quad

20. apple_pi

Here's an example you might want to look at: tan(4x) = -tan(2x) 0° ≤ x ≤ 360° tan(4x) = tan(-2x) ****DONT DO THIS: 4x = -2x INSTEAD... 4x = 180n -2x (Write one side as a general solution) then simplify: 6x = 180n x = 30n then you have all possible values of x now find all the ones that work, and you have all solutions