# How do you solve tanx/cosx=2?

Jul 28, 2016

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

#### Explanation:

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

$\tan \frac{x}{\cos} x = \frac{\sin \frac{x}{\cos} x}{\cos} x = \sin \frac{x}{\cos} ^ 2 x$

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

$\sin \frac{x}{1 - {\sin}^{2} x} = 2$

It is equivalent to:

$\sin x = 2 \left(1 - {\sin}^{2} x\right) \mathmr{and} \cos x \ne 0$

$2 {\sin}^{2} x + \sin x - 2 = 0 \mathmr{and} x \ne \frac{\pi}{2} + k \pi$

$\sin x = \frac{- 1 \pm \sqrt{1 + 16}}{4}$

• Case when $\sin x < - 1$ can be dropped:

the solution

$\sin x = \frac{- 1 - \sqrt{17}}{4} < - 1$

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

• The solution is

$\sin x = \frac{- 1 + \sqrt{17}}{4}$

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