Ask your own question, for FREE!
Mathematics 19 Online
OpenStudy (anonymous):

solve tan x sec x - 2 tan x = 0 for all real values of x..

OpenStudy (anonymous):

@jim_thompson5910

OpenStudy (anonymous):

@dan815

OpenStudy (rajat97):

sec x=2 is it the solution?

OpenStudy (anonymous):

i wish it was that easy but its not :'( the answer has multiple solutions

jimthompson5910 (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

OpenStudy (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 ?

OpenStudy (rajat97):

got it

OpenStudy (rajat97):

the kpi solution is for sin x=0 and the other two ae for sec x=2

OpenStudy (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

OpenStudy (usukidoll):

depends on the quadrant you're in as well .

OpenStudy (anonymous):

okay so the answers i have to select from are sketchy so i think i picked the right one? ..

jimthompson5910 (jim_thompson5910):

they made a typo they should have 2k*pi and not just 2k

jimthompson5910 (jim_thompson5910):

when you wrote `0+kpi ` `2kpi + pi/3` `2kpi + 5pi/3` you have all 3 correct

jimthompson5910 (jim_thompson5910):

5pi/3 is coterminal to -pi/3, so either works

OpenStudy (anonymous):

so judging on which one is the closest... it'd be A?

jimthompson5910 (jim_thompson5910):

you'll have to bring it up with your teacher so they can fix it

jimthompson5910 (jim_thompson5910):

but yeah A looks closest

OpenStudy (anonymous):

i will thanks jim you da bestttttttttttttttt

jimthompson5910 (jim_thompson5910):

yw

OpenStudy (rajat97):

when you solve it, you get sin x=0 and sec x=2 so you get the answer as given

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!