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

Question

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

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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

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