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

User LK Yeung
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1 Answer

3 votes

Answer:


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

User EinsA
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