How to solve #sin(x+pi/6)# = #2sin(x-pi/6)# in radian form?

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#sin(x+pi/6)#
#2sin(x-pi/6)#

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Answer 1 · 2016-10-12T16:12:14

#x = kpi+pi/6, k = 0, +-1, +-2, +-3, ...#

Rearranging,

#sin(x+pi/6)-sin(x-pi/6)=sin(x-pi/6)#. So,

#2 cos x sin (pi/6)=sin x cos (pi/6)-cos x sin (pi/6)#. Simplifying,

#cos x = 0 to cos (pi/6) = 0#. So, cos x is not 0.

# 3 cos x sin (pi/5)=sin x cos (pi/6)#. Further simplification leads to

#tan x= 3 tan (pi/6) = 3/sqrt 3 = 1/sqrt 3 = tan (pi/6)#.

So, the general value of x is

#x = kpi+pi/6, k = 0, +-1, +-2, +-3, ...#