britbrat4290
Solve the equation on the interval [0, 2π).
sin2 x - cos2 x = 0
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britbrat4290
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sry no lol
Coolsector
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and you should convince yourself why it is fine to divide by cos2x . ( it is fine. )
britbrat4290
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wud it b pie/4, pie/6
?
Coolsector
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wait im stupid
Coolsector
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tan(2x) = 1 !!
britbrat4290
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??
Coolsector
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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
britbrat4290
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pie/4 just then
Coolsector
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no
britbrat4290
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u got over 8?
Coolsector
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pi/8 , 5pi/8 , 9pi/8 , 13pi/8
Coolsector
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i first wrote tan(2x) = 0 which was WRONG
britbrat4290
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pi/4, 3pi/4, 5pi/4, 7pi/4
Coolsector
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no,
x=pi/8 , 5pi/8 , 9pi/8 , 13pi/8
britbrat4290
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hmm
Coolsector
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wait lol
was it :
Coolsector
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\[\sin(2x) - \cos(2x) = 0\]
Coolsector
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or
\[\sin^2(x) - \cos^2(x) = 0\]
Coolsector
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cause i solved for the FIRST one
Coolsector
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?
britbrat4290
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it said i got it right for the four i picked? weird
Coolsector
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but what was the correct question
i wrote here two
Coolsector
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i think i solved for the wrong one
britbrat4290
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im lost sry its ok bc i got it right so its all good :)
Coolsector
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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
Coolsector
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and i thought it was sin(2x) - cos(2x) = 0