How do you solve #sinx cos (pi/6) + sin (pi/6) cos x= 1/sqrt2#?

1 Answer
Nghi N.
Apr 12, 2016

#(pi)/12 and (7pi)/12#

Explanation:

Use the trig identity:
sin a.cos b + sin b.cos a = sin (a + b)
#sin x.cos (pi/6) + sin (pi/6).cos x = sin (x + pi/6).#
The equation transforms to:
#sin (x + pi/6) = 1/sqrt2 = sin (pi/4)#.
The trig unit circle gives another arc #(3pi)/4# that has the same sin value #(1/sqrt2).#
a. #x + pi/6 = pi/4# --> #x = pi/4 - pi/6 = pi/12#
b. #x + pi/6 = (3pi)/4# ---> #x = (3pi)/4 - pi/6 = (7pi)/12#
Answer for #(0, 2pi)#:
#pi/12 and (7pi)/12#

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