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3psilon
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sinx + 2cosx = 0 where 0<=x<=2pi, what can x be equal to
 one year ago
 one year ago
3psilon Group Title
sinx + 2cosx = 0 where 0<=x<=2pi, what can x be equal to
 one year ago
 one year ago

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nightwill Group TitleBest ResponseYou've already chosen the best response.1
sinx=2cosx sinx/cosx=2 tanx=2 Whatever x that fits tanx=2
 one year ago

3psilon Group TitleBest ResponseYou've already chosen the best response.0
so I did \[\tan^{1}(2)\] and I got 1.107 which isn't correct
 one year ago

3psilon Group TitleBest ResponseYou've already chosen the best response.0
@nightwill
 one year ago

nightwill Group TitleBest ResponseYou've already chosen the best response.1
x has to be between 0 and 2pi, so 1.107 won't work.
 one year ago

nightwill Group TitleBest ResponseYou've already chosen the best response.1
the answer should be pi+arctan(2) which is about 2.03444
 one year ago

3psilon Group TitleBest ResponseYou've already chosen the best response.0
Why did you add pi ?
 one year ago

3psilon Group TitleBest ResponseYou've already chosen the best response.0
I'm trying to refresh on trig
 one year ago

nightwill Group TitleBest ResponseYou've already chosen the best response.1
I added pi so that it fits in the domain. tan x has a period of pi, so tan x = tan (x+pi) Therefore, \[\tan(tan^{1}2)=tan(\pi+tan^{1}2)\]
 one year ago
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