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

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britbrat4290 Group TitleBest ResponseYou've already chosen the best response.0
sry no lol
 one year 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. )
 one year ago

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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