Question

how to solve tan(x+π)+2sin(x+π)=0

Answer 1

x+pi for tan translates to tan of x

x+pi for sine translates to sine of -x.

so tan(x+π)+2sin(x+π)=0

= tan x - 2 sin x = 0

tan x = sinx/cos x

sin x - 2sin x cos x = 0
sin x (1-2cos x) = 0
sin x =0, 1-2cos x =0

Solve them for the general solutions!

Got it? :)

Answer 2

yes thanks

Answer 3

since tan(x+y)=tanx+tany\1-tanxtany and sin(a+b)=sinacosb+cosasinb

\[\frac{ tanx+\tan \pi }{ 1-tanxtan \pi }+2[sinxcos \pi+cosxsin \pi]\]=0
\[2\pi=360 so \pi=180\]
tan180=0 , sin180=0 but cos180=-1
now put it in the given equation
\[\frac{ tanx +0 }{ 1- tanx(0) }+2[sinx(-1)+cosx(0)]\]=0

\[tanx -2sinx =0\]
\[\frac{ sinx }{ cosx }-2sinx=0\]
sinx(1\cosx -2)=0
sinx(secx-2)=0
since,1\cosx=secx
either,
sinx=0 or secx+2=0
x=sin^-1(0) or secx=-2
x=0 or x=sec^-1(2)
x=0 or x= 120
x=0 or x=2pi\3
this is the correct answer