anonymous
  • anonymous
solve tan x sec x - 2 tan x = 0 for all real values of x..
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
anonymous
  • anonymous
rajat97
  • rajat97
sec x=2 is it the solution?

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anonymous
  • anonymous
i wish it was that easy but its not :'( the answer has multiple solutions
jim_thompson5910
  • jim_thompson5910
tan(x)*sec(x) - 2*tan(x) = 0 tan(x)*[ sec(x) - 2 ] = 0 tan(x) = 0 or sec(x) - 2 = 0 tan(x) = 0 or sec(x) = 2 tan(x) = 0 or 1/cos(x) = 2 tan(x) = 0 or cos(x) = 1/2 I'll let you finish up
anonymous
  • anonymous
@jim_thompson5910 okay i understand.. so with that being said.. they're will be 3 different solutions, and they would be 0+kpi 2kpi + pi/3 2kpi + 5pi/3 ?
rajat97
  • rajat97
got it
rajat97
  • rajat97
the kpi solution is for sin x=0 and the other two ae for sec x=2
UsukiDoll
  • UsukiDoll
it's been a while since I've done this but you have to be in radian mode and add or subtract 2pi or something
UsukiDoll
  • UsukiDoll
depends on the quadrant you're in as well .
anonymous
  • anonymous
okay so the answers i have to select from are sketchy so i think i picked the right one? ..
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jim_thompson5910
  • jim_thompson5910
they made a typo they should have 2k*pi and not just 2k
jim_thompson5910
  • jim_thompson5910
when you wrote `0+kpi ` `2kpi + pi/3` `2kpi + 5pi/3` you have all 3 correct
jim_thompson5910
  • jim_thompson5910
5pi/3 is coterminal to -pi/3, so either works
anonymous
  • anonymous
so judging on which one is the closest... it'd be A?
jim_thompson5910
  • jim_thompson5910
you'll have to bring it up with your teacher so they can fix it
jim_thompson5910
  • jim_thompson5910
but yeah A looks closest
anonymous
  • anonymous
i will thanks jim you da bestttttttttttttttt
jim_thompson5910
  • jim_thompson5910
yw
rajat97
  • rajat97
when you solve it, you get sin x=0 and sec x=2 so you get the answer as given

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