Question

Tan(x+pie) + sin(x+pie) + sin (x-pie) = 0

Answer 1

`x=(\pi )/(3)+2\pi n, \quad x=(5\pi )/(3)+2\pi n,\quad x=2\pi n,\quad x=\pi+2\pi n`

Explanation:

`\boxed{\begin{minipage}{6.5 cm}\underline{Trigonometric Double Angle Identities}\\\\$\sin (A \pm B)=\sin A \cos B \pm \cos A \sin B$\\\\$\cos (A \pm B)=\cos A \cos B \mp \sin A \sin B$\\\\$\tan (A \pm B)=(\tan A \pm \tan B)/(1 \mp \tan A \tan B)$\\\end{minipage}}`

Given equation:

`\tan (x + \pi) + \sin(x + \pi) + \sin(x - \pi)=0`

Simplify using the double angle identities:

`\implies (\tan x + \tan \pi)/(1 -\tan x \tan \pi) + \sin x \cos \pi+\cos x \sin \pi + \sin x \cos \pi -\cos x \sin \pi=0`

`\implies (\tan x + 0)/(1 -\tan x(0)) + \sin x (-1)+\cos x (0) + \sin x(-1) -\cos x (0)=0`

`\implies (\tan x)/(1) -\sin x - \sin x=0`

`\implies \tan x-2\sin x =0`

Use the tan identity to rewrite tan(x) in terms of sin(x) and cos(x):

`\implies (\sin x)/( \cos x)-2\sin x=0`

Multiply both sides by cos(x):

`\implies (\sin x\cos x)/( \cos x)-2\sin x\cos x=0(\cos x)`

`\implies \sin x-2\sin x \cos x=0`

Factor out sin(x) from the left side of the equation:

`\implies \sin x(1-2 \cos x)=0`

Apply the zero-product property.

Case 1

`\implies \sin x=0`

`\implies x=\sin^(-1)(0)`

`\implies x=2\pi n, \; \pi+2\pi n`

Case 2

`\implies 1-2 \cos x=0`

`\implies 1=2 \cos x`

`\implies \cos x=(1)/(2)`

`\implies x= \cos ^(-1) \left((1)/(2)\right)`

`\implies x=(\pi )/(3)+2\pi n, \;(5\pi )/(3)+2\pi n`

Solution

`x=(\pi )/(3)+2\pi n, \quad x=(5\pi )/(3)+2\pi n,\quad x=2\pi n,\quad x=\pi+2\pi n`