How do you solve #tanx/cosx=2#?

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Answer 1 · 2016-07-28T15:19:07

#x=sin^-1((-1+sqrt(17))/4)#

Since #tan x=sinx/cosx#, you can substitute:

#tanx/cosx=(sinx/cosx)/cosx=sinx/cos^2x#

Since #cos^2x=1-sin^2x#, the given equation becomes:

#sinx/(1-sin^2x)=2#

It is equivalent to:

#sinx=2(1-sin^2x) and cosx!=0#

#2sin^2x+sinx-2=0 and x!=pi/2+kpi#

#sinx=(-1+-sqrt(1+16))/4#

the solution

#sinx=(-1-sqrt(17))/4<-1#

the range of sin x is [-1;1] therefore

#sinx=(-1+sqrt(17))/4#

#x=sin^-1((-1+sqrt(17))/4)#