How do you solve #\sin 4\theta = \frac { \sqrt { 3} } { 2}#?

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Answer 1 · 2017-04-22T22:04:17

#pi/12 + (kpi)/2#
#pi/6 + (kpi)/2#

#sin 4t = sqrt3/2#
Trig table and unit circle give 2 solutions:

a. #4t = pi/3 + 2kpi#
#t = pi/12 + (kpi)/2#
b. #4t = (2pi)/3 + 2kpi#
#t = pi/6 + (kpi)/2#

Answer 2 · 2017-04-23T08:33:45

#sin4theta = sqrt(3)/2#

so, #4theta = pi/3 + 2npi# or #(pi - pi/3) + 2npi# for #n in ZZ#

i.e. #4theta = pi/3 + 2npi# or #(2pi)/3 + 2npi#

#=> theta = pi/12 + (npi)/2 or pi/6 + (npi)/2#

or, #theta = (6n + 1)pi/12 or (1 + 3n)pi/6#

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