## britbrat4290 2 years ago Solve the equation on the interval [0, 2π). sin2 x - cos2 x = 0

1. britbrat4290

sry no lol

2. Coolsector

and you should convince yourself why it is fine to divide by cos2x . ( it is fine. )

3. britbrat4290

wud it b pie/4, pie/6 ?

4. Coolsector

wait im stupid

5. Coolsector

tan(2x) = 1 !!

6. britbrat4290

??

7. Coolsector

so the general solution for tan(x) = tan(y) x = y + pi*k so in our case: tan(2x) = tan(pi/4) 2x = pi/4 + pi*k x = pi/8 + (pi/2) * k

8. britbrat4290

pie/4 just then

9. Coolsector

no

10. britbrat4290

u got over 8?

11. Coolsector

pi/8 , 5pi/8 , 9pi/8 , 13pi/8

12. Coolsector

i first wrote tan(2x) = 0 which was WRONG

13. britbrat4290

pi/4, 3pi/4, 5pi/4, 7pi/4

14. Coolsector

no, x=pi/8 , 5pi/8 , 9pi/8 , 13pi/8

15. britbrat4290

hmm

16. Coolsector

wait lol was it :

17. Coolsector

$\sin(2x) - \cos(2x) = 0$

18. Coolsector

or $\sin^2(x) - \cos^2(x) = 0$

19. Coolsector

cause i solved for the FIRST one

20. Coolsector

?

21. britbrat4290

it said i got it right for the four i picked? weird

22. Coolsector

but what was the correct question i wrote here two

23. Coolsector

i think i solved for the wrong one

24. britbrat4290

im lost sry its ok bc i got it right so its all good :)

25. Coolsector

cause if the question was : sin^2(x) - cos^2(x) = 0 then we get tan^2(x) = 1 tan(x) = +/-1 so 1. tan(x) = 1 tan(x) = tan(pi/4) x = pi/4 + pi * k x = pi/4 , 5pi/4 2.tan(x) = -1 tan(x) = tan(-pi/4) x = -pi/4 + pi*k x = 3pi/4 , 7pi/4 so : x = pi/4, 3pi/4, 5pi/4, 7pi/4

26. Coolsector

and i thought it was sin(2x) - cos(2x) = 0