# What is the equation of the line that is normal to f(x)= tanx-sin2x at  x=(4pi)/3 ?

$\textcolor{b l u e}{y = - \frac{1}{5} x + \frac{4 \pi}{15} + \frac{\sqrt{3}}{2}}$

#### Explanation:

Given

$f \left(x\right) = \tan x - \sin 2 x$ at ${x}_{1} = \frac{4 \pi}{3}$

Solve for the point $\left({x}_{1} , {y}_{1}\right)$ first

${y}_{1} = \tan {x}_{1} - \sin 2 \cdot {x}_{1}$

${y}_{1} = \tan \left(\frac{4 \pi}{3}\right) - \sin \left(2 \cdot \frac{4 \pi}{3}\right)$

${y}_{1} = \sqrt{3} - \frac{\sqrt{3}}{2}$

${y}_{1} = \frac{\sqrt{3}}{2}$

Our point $\left({x}_{1} , {y}_{1}\right) = \left(\frac{4 \pi}{3} , \frac{\sqrt{3}}{2}\right)$

Solve for the slope $m$

$f \left(x\right) = \tan x - \sin 2 x$

find the first derivative $f ' \left(x\right) = m$

$f ' \left(x\right) = {\sec}^{2} x - 2 \cdot \cos 2 x$

$m = f ' \left(\frac{4 \pi}{3}\right) = {\sec}^{2} \left(\frac{4 \pi}{3}\right) - 2 \cdot \cos \left(2 \left(\frac{4 \pi}{3}\right)\right)$
$m = {\left(- 2\right)}^{2} - 2 \left(- \frac{1}{2}\right)$
$m = 4 + 1 = 5$

For the normal line

${m}_{n} = - \frac{1}{m}$

${m}_{n} = - \frac{1}{5}$

Solve for the normal line:

$y - {y}_{1} = {m}_{n} \left(x - {x}_{1}\right)$

$y - \frac{\sqrt{3}}{2} = - \frac{1}{5} \left(x - \frac{4 \pi}{3}\right)$

$y - \frac{\sqrt{3}}{2} = - \frac{1}{5} x + \frac{4 \pi}{15}$

$\textcolor{b l u e}{y = - \frac{1}{5} x + \frac{4 \pi}{15} + \frac{\sqrt{3}}{2}}$

Kindly see the graph of $f \left(x\right) = \tan x - \sin 2 x$ and the normal line $y = - \frac{1}{5} x + \frac{4 \pi}{15} + \frac{\sqrt{3}}{2}$ at the point $\left(\frac{4 \pi}{3} , \frac{\sqrt{3}}{2}\right)$

God bless....I hope the explanation is useful.