How do you solve for x in #sin(210+x)-cos(120+x)=0#?

1 Answer
Zor Shekhtman
Feb 18, 2015

We assume that #210# and #120# and angles in degrees.
This equation can be transformed into identity #0=0#, which means that any real value of #x# is a valid solution. Here is why.

Let's use the known equalities that immediately follow from the definition of #sin(phi)# and #cos(phi)# for any angle #phi# using a unit circle (all angles are measured in degrees):
#sin(180+phi)=-sin(phi)#
#cos(90+phi)=-sin(phi)#

Applying these equalities to our equation, we obtain
#sin(210+x)−cos(120+x)=0#
#sin(180+30+x)−cos(90+30+x)=0#
#-sin(30+x)+sin(30+x)=0#
#0=0#
The fact that we came to an unconditional identity for any variable #x# signifies that any real value of #x# is a solution.

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