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

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

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

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