How do you find all solutions of the equation #tan(x+pi)+2sin(x+pi)=0# in the interval #[0,2pi)#?

1 Answer
Somebody N.
Nov 9, 2017

#color(blue)(0 , pi , pi/3 , (11pi)/6)#

Explanation:

We can use the identities:

#color(red)(sin(A+B)= sinAcosB+cosAsinB)#

#color(red)(cos(A+B)=cosAcosB-sinAsinB)#

#color(red)(tanA=sinA/cosB)#
..................................................................................................................................

#tan(x+pi)=(sin(x)cospi+cosxsinpi)/(cosxcospi-sinxsinpi)#

#(sin(x)cospi+cosxsinpi)/(cosxcospi-sinxsinpi)+2(sinxcospi+cosxsinpi)=0#

#(sin(x)(-1)+cosx(0))/(cosx(-1)-sinx(0))+2(sinx(-1)+cosx(0))=0#

#(-sin(x))/(-cosx)-2sinx=0#

#sinx-2sinxcosx=0#

#sinx(1-2cosx)=0#

#sinx= 0=>x= color(blue)(0 , pi) #

#cosx=1/2=>x= color(blue)(pi/3 , (5pi)/3)#

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