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

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sry no lol
and you should convince yourself why it is fine to divide by cos2x . ( it is fine. )
wud it b pie/4, pie/6 ?

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Other answers:

wait im stupid
tan(2x) = 1 !!
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
pie/4 just then
u got over 8?
pi/8 , 5pi/8 , 9pi/8 , 13pi/8
i first wrote tan(2x) = 0 which was WRONG
pi/4, 3pi/4, 5pi/4, 7pi/4
no, x=pi/8 , 5pi/8 , 9pi/8 , 13pi/8
wait lol was it :
\[\sin(2x) - \cos(2x) = 0\]
or \[\sin^2(x) - \cos^2(x) = 0\]
cause i solved for the FIRST one
it said i got it right for the four i picked? weird
but what was the correct question i wrote here two
i think i solved for the wrong one
im lost sry its ok bc i got it right so its all good :)
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
and i thought it was sin(2x) - cos(2x) = 0

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