If #sin(60-x)-sin(60-x)=0# what is #x#?

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Answer 1 · 2017-02-22T14:36:53

#x=kpi, k in ZZ#

#sin(60-x)-sin(60-x)=0#
So
#sin(60-x)-sin(60-x)=-sin(x)#
#rarr -sin(x)=0#
#rarr sin(x)=0# which is true for #x=0# and all integer multiples of #pi#.

Answer 2 · 2017-02-22T14:42:13

Perhaps you mean #sin (60-x)-sin(60+x)=-sinx#. Please see below for solution.

Perhaps you mean #sin (60-x)-sin(60+x)=-sinx#

As #sinC-sinD=2cos((C+D)/2)sin((C-D)/2)#

Hence #sin (60-x)-sin(60+x)=2cos((60-x+60+x)/2)sin((60-x-60-x)/2)#

= #2cos60sin(-x)#

= #2xx1/2xx(-sinx)#

= #-sinx#