How do you solve $\frac{tan x}{cos x} = 2$ $tanx/cosx=2$ ?

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Answer 1 · 2016-07-28T15:19:07

$x = {sin}^{- 1} (\frac{- 1 + \sqrt{17}}{4})$ $x=sin^-1((-1+sqrt(17))/4)$

Since $tan x = \frac{sin x}{cos x}$ $tan x=sinx/cosx$ , you can substitute:

$\frac{tan x}{cos x} = \frac{\frac{sin x}{cos x}}{cos x} = \frac{sin x}{{cos}^{2} x}$ $tanx/cosx=(sinx/cosx)/cosx=sinx/cos^2x$

Since ${cos}^{2} x = 1 - {sin}^{2} x$ $cos^2x=1-sin^2x$ , the given equation becomes:

$\frac{sin x}{1 - {sin}^{2} x} = 2$ $sinx/(1-sin^2x)=2$

It is equivalent to:

$sin x = 2 (1 - {sin}^{2} x) and cos x \neq 0$ $sinx=2(1-sin^2x) and cosx!=0$

$2 {sin}^{2} x + sin x - 2 = 0 and x \neq \frac{π}{2} + k π$ $2sin^2x+sinx-2=0 and x!=pi/2+kpi$

$sin x = \frac{- 1 \pm \sqrt{1 + 16}}{4}$ $sinx=(-1+-sqrt(1+16))/4$

the solution

$sin x = \frac{- 1 - \sqrt{17}}{4} < - 1$ $sinx=(-1-sqrt(17))/4<-1$

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

$sin x = \frac{- 1 + \sqrt{17}}{4}$ $sinx=(-1+sqrt(17))/4$

$x = {sin}^{- 1} (\frac{- 1 + \sqrt{17}}{4})$ $x=sin^-1((-1+sqrt(17))/4)$

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