How do you solve $sin(x+pi/6)-sin(x-pi/6)=1/2$ ?

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Answer 1 · 2015-05-15T19:32:50

Use the trig identity: sin (a + b) - sin (a - b) = 2cos a.sin b.

Left side:

$sin (x + pi/6) - sin (x - pi/6) = 2 cos x.sin (pi/6) = 2cos x(1/2) = cos x$

Finally: $cos x = 1/2$ -> $x = pi/3 and x = -pi/3$

Within period (0, 2pi), there are 2 answers; $pi/3 and -pi/3 or ((5pi)/6)$

How do you solve sin(x+pi/6)-sin(x-pi/6)=1/2sin(x+pi/6)-sin(x-pi/6)=1/2?