# How do you solve tan(x+pi)+cos(x+pi/2)=0?

Jul 24, 2016

This can be solved as follows:

#### Explanation:

$\tan \left(\pi + x\right) + \cos \left(\pi + x\right) = 0$
$\tan x + \left(- \sin x\right) = 0$ $\left(e q u a t i o n .1\right)$

$\sin \frac{x}{\cos} x - \sin x = 0$

$\left(\sin x - \sin x \cos x\right) = 0$

$\sin x \frac{1 - \cos x}{\cos} x = 0$

$\sin x \left(1 - \cos x\right) = 0$

$1 - \cos x = 0$

$1 = \cos x$

$\cos 0 = \cos x$

$\therefore , x = 0$

$p u t$ $t h e$ $v a l u e$ $o f x$ in $e q .1$

$\tan 0 - \sin 0$
$= 0$
$= R . H . S$

Jul 24, 2016

Solution is $x = n \pi$
where $n$ is an integer.

#### Explanation:

$\tan \left(x + \pi\right) + \cos \left(x + \frac{\pi}{2}\right) = 0$
Using the identities $\tan \left(\pi + A\right) = \tan A \mathmr{and} \cos \left(B + \frac{\pi}{2}\right) = - \sin B$, we get

$\tan x + \left(- \sin x\right) = 0$ ...............(1)

$\implies \sin \frac{x}{\cos} x - \sin x = 0$

$\implies \frac{\sin x - \sin x \cos x}{\cos} x = 0$

Muliplying both sides with $\cos x$ we get

$\sin x \left(1 - \cos x\right) = 0$

Setting both factors equal to zero
$\sin x = 0$ .......(2)
$1 - \cos x = 0$ ......(3)

From (2), Due to sinusoidal nature of both $\sin \mathmr{and} \cos$ functions we get:

$x = n \pi$ .....(4)
where $n$ is an integer.
graph{y=sin x [-20, 20, -2, 2]}

Similarly from (3) we get

$\cos x = 1$
graph{y=cos x [-20, 20, -2, 2]}

$x = n \left(2 \pi\right)$ .....(5)

From (4) and (5)
$x = n \pi$