How do you solve #\sin 2x + \cos ( - x ) = 0#?

It helps to know what range that #x# could fall into for trigonometric equations. Otherwise, the sinusoidal nature of the equation will have an infinite number of solutions! Let's assume that you want to find the solutions on the interval #[0, 2pi]#.

Cosine is an even function, so rewrite as
#sin(2x)+cos(x)=0#

Isolate cosine to the left-hand side. Subtract #sin(2x)# from both sides:
#cos(x)=-sin(2x)#

Express the right hand side in terms of cosine. Rewrite the right-hand side using #-sin(theta)=cos(-theta-pi//2)#

#cos(x)=cos(-2x-pi/2)#

Taking the inverse cosine of both sides reveals that the arguments are equal to each other.

#cos^-1[cos(x)]=cos^-1[cos(-2x-pi/2)]#
#x=-2x-pi/2+2pin# for #ninZZ#

Solve for #x# by adding #2x# to both sides
#3x=-pi/2+2pin#
#x=-pi/6+2/3pin#

In the interval #[0, 2pi]#, the solutions are
#x=pi/2, 7pi/6,# and #11pi/6#

Solve #sin2x+cos(-x)=0#

Consider the following identities:

Double angle identity for sine:
#sin2x=2sinxcosx#

Even-odd identity for cosine:
#cos(-x)=cos(x)#

We can substitute these identities into the original equation:
#2sinxcosx+cosx=0#

We can factor out a #cosx#:
#cosx(2sinx+1)=0#

So we have that #cosx=0# or #2sinx+1=0#, which we can solve individually:
#cosx=0#
#x=pi/2+npi# where #n in RR#

#2sinx+1=0#
#sinx=-1/2#
#x=(7pi)/6+2pin# where #n in RR#
#x=(11pi)/6+2pin# where #n in RR#

#therefore# #x=pi/2+npi,(7pi)/6+2pin,(11pi)/6+2pin# where #n in RR#