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britbrat4290
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Solve the equation on the interval [0, 2π).
sin2 x  cos2 x = 0
 11 months ago
 11 months ago
britbrat4290 Group Title
Solve the equation on the interval [0, 2π). sin2 x  cos2 x = 0
 11 months ago
 11 months ago

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britbrat4290 Group TitleBest ResponseYou've already chosen the best response.0
sry no lol
 11 months ago

Coolsector Group TitleBest ResponseYou've already chosen the best response.0
and you should convince yourself why it is fine to divide by cos2x . ( it is fine. )
 11 months ago

britbrat4290 Group TitleBest ResponseYou've already chosen the best response.0
wud it b pie/4, pie/6 ?
 11 months ago

Coolsector Group TitleBest ResponseYou've already chosen the best response.0
wait im stupid
 11 months ago

Coolsector Group TitleBest ResponseYou've already chosen the best response.0
tan(2x) = 1 !!
 11 months ago

Coolsector Group TitleBest ResponseYou've already chosen the best response.0
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
 11 months ago

britbrat4290 Group TitleBest ResponseYou've already chosen the best response.0
pie/4 just then
 11 months ago

britbrat4290 Group TitleBest ResponseYou've already chosen the best response.0
u got over 8?
 11 months ago

Coolsector Group TitleBest ResponseYou've already chosen the best response.0
pi/8 , 5pi/8 , 9pi/8 , 13pi/8
 11 months ago

Coolsector Group TitleBest ResponseYou've already chosen the best response.0
i first wrote tan(2x) = 0 which was WRONG
 11 months ago

britbrat4290 Group TitleBest ResponseYou've already chosen the best response.0
pi/4, 3pi/4, 5pi/4, 7pi/4
 11 months ago

Coolsector Group TitleBest ResponseYou've already chosen the best response.0
no, x=pi/8 , 5pi/8 , 9pi/8 , 13pi/8
 11 months ago

Coolsector Group TitleBest ResponseYou've already chosen the best response.0
wait lol was it :
 11 months ago

Coolsector Group TitleBest ResponseYou've already chosen the best response.0
\[\sin(2x)  \cos(2x) = 0\]
 11 months ago

Coolsector Group TitleBest ResponseYou've already chosen the best response.0
or \[\sin^2(x)  \cos^2(x) = 0\]
 11 months ago

Coolsector Group TitleBest ResponseYou've already chosen the best response.0
cause i solved for the FIRST one
 11 months ago

britbrat4290 Group TitleBest ResponseYou've already chosen the best response.0
it said i got it right for the four i picked? weird
 11 months ago

Coolsector Group TitleBest ResponseYou've already chosen the best response.0
but what was the correct question i wrote here two
 11 months ago

Coolsector Group TitleBest ResponseYou've already chosen the best response.0
i think i solved for the wrong one
 11 months ago

britbrat4290 Group TitleBest ResponseYou've already chosen the best response.0
im lost sry its ok bc i got it right so its all good :)
 11 months ago

Coolsector Group TitleBest ResponseYou've already chosen the best response.0
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
 11 months ago

Coolsector Group TitleBest ResponseYou've already chosen the best response.0
and i thought it was sin(2x)  cos(2x) = 0
 11 months ago
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