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

Question

7842 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-24T07:20:12

This can be solved as follows:

#tan(pi+x) + cos(pi+x)=0#
#tanx + (-sinx)=0# #(equation.1)#

#sinx/cosx -sinx=0#

#(sinx-sinxcosx)=0#

#sinx(1-cosx)/cosx=0#

#sinx(1-cosx)=0#

#1-cosx=0#

#1=cosx#

#cos0=cosx#

#therefore,x=0#

#put# #the# #value# #of x# in #eq.1#

#tan0 - sin0 #
#=0#
#=R.H.S#

Answer 2 · 2016-07-24T08:40:29

Solution is #x=npi#
where #n# is an integer.

#tan(x+pi) + cos(x+pi/2)=0#
Using the identities #tan(pi+A)=tanA and cos(B+pi/2)=-sinB#, we get

#tanx + (-sinx)=0# ...............(1)

#=>sinx/cosx -sinx=0#

#=>(sinx-sinxcosx)/cosx=0#

Muliplying both sides with #cosx# we get

#sinx(1-cosx)=0#

Setting both factors equal to zero
#sinx=0# .......(2)
#1-cosx=0# ......(3)

From (2), Due to sinusoidal nature of both #sin and cos# functions we get:

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

Similarly from (3) we get

#cosx=1#
graph{y=cos x [-20, 20, -2, 2]}

#x=n (2pi)# .....(5)

From (4) and (5)
#x=npi#