How do you solve #sin theta= -0.5 #?

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Answer 1 · 2018-05-19T02:04:53

#(7pi)/6 + pi * k#, where k is all integers.

If they are asking the value on #theta# on the interval #0 <= theta <= 2pi# the solutions are: #(7pi)/6 and (11pi)/6#

#sin(theta) = -0.5#
#sin(theta) = -1/2#

Looking at the unit circle, #sin(theta) = -1/2# at #(7pi)/6 and (11pi)/6# but since sine is periodic every #2pi# the solutions are:

#(7pi)/6 + pi * k#, where k is all integers.

Answer 2 · 2018-05-19T02:16:53

#theta = 2n pi - pi/6, 2npi - (5pi)/6#, n #1 in ZZ#

#sin theta = -0.5#

#sin theta = sin (-pi/6)#

#theta = -pi/6#

Generalising

#theta = 2n pi - pi/6, 2npi - (5pi)/6#, n #1 in ZZ#