Question

Solve the equation on the interval [0, 2π).

sin2 x - cos2 x = 0

Answer 1

sry no lol

Answer 2

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

Answer 3

wud it b pie/4, pie/6
?

Answer 4

wait im stupid

Answer 5

tan(2x) = 1 !!

Answer 6

??

Answer 7

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

Answer 8

pie/4 just then

Answer 9

no

Answer 10

u got over 8?

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

no,

x=pi/8 , 5pi/8 , 9pi/8 , 13pi/8

Answer 15

hmm

Answer 16

wait lol
was it :

Answer 17

\[\sin(2x) - \cos(2x) = 0\]

Answer 18

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

Answer 19

cause i solved for the FIRST one

Answer 20

?

Answer 21

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

Answer 22

but what was the correct question
i wrote here two

Answer 23

i think i solved for the wrong one

Answer 24

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

Answer 25

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

Answer 26

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