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

2 Answers
Jul 24, 2016

This can be solved as follows:

Explanation:

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

Jul 24, 2016

Solution is x=npi
where n is an integer.

Explanation:

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