## jmartinez638 one year ago How do you find all the solutions for 'sin(x/2 - pi/4) = sqrt2/2'?

1. phi

solutions imply an equation equations have an = sign your expression does not. is that a typo?

2. jmartinez638

Yes, So sorry = sqrt2/2

3. jmartinez638

Edited original.

4. phi

do you know how to "undo" sin ?

5. jmartinez638

Like using identities?

6. phi

more like, do $$\sin^{-1}$$ to both sides

7. phi

$\sin^{-1} \left(\sin\left(\frac{x}{2} - \frac{\pi}{4}\right) \right)=\sin^{-1} \frac{\sqrt2}{2}$

8. phi

on the left side, the inverse sin of the sin undoes the sin we are left with $\frac{x}{2} - \frac{\pi}{4}=\sin^{-1} \frac{\sqrt2}{2}$

9. phi

on the right side it is asking for an angle what angle is it where sin of that angle = sqr(2)/2 we want it in radians and it is an angle people memorize (so you should too)

10. jmartinez638

45 degrees or $\pi$/4

11. phi

yes. but because the question asks for *all* solutions we should eyeball the graph for sin |dw:1440367395779:dw|

12. phi

so one family of solutions is pi/4 + 2pi n and the other is 3pi/4 + 2pi n where n is any integer

13. jmartinez638

Oh that makes sense. In relation to the graph especially...

14. phi

so we should try to solve $\frac{x}{2} - \frac{\pi}{4}=\frac{\pi}{4} + 2\pi n$ and also $\frac{x}{2} - \frac{\pi}{4}=\frac{3\pi}{4} + 2\pi n$

15. jmartinez638

$(8n+5)\pi \div2$ for the first one

16. jmartinez638

17. phi

$\frac{x}{2} - \frac{\pi}{4}=\frac{\pi}{4} + 2\pi n \\ \frac{x}{2} = \frac{\pi}{2} + 2\pi n \\ x= \pi + 4 \pi n$

18. jmartinez638

Oops yeah, I must've looked at the second one and not the first :3 Let me try the first...

19. jmartinez638

I got x=(4n+1)*pi

20. phi

yes , that looks good

21. phi

and for the 2nd equation?

22. jmartinez638

Does (8n+5)π÷2 work?

23. phi

I don't see how.

24. jmartinez638

Ah, I seem to have made a mistake. How about $x=2(2n+1)\pi$?

25. phi

yes, that looks good. though people would probably write it as 2pi(2n+1)

26. phi

and of course we have to say n is any integer

27. jmartinez638

Ok, of course. So those two equations, with 'n' designated as any integer, would be the solutions to that equation?

28. phi

yes

29. jmartinez638

That makes a lot of sense, thank you so much!

30. phi

yw