How do you solve #sin3x=1/2#?

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Answer 1 · 2015-06-09T07:17:31

From angle table

You know that

#sin(30^@)=1/2#

In your equation, you have

#sin(3x)=1/2#

Since both equations are equal to the same value, you can write

#sin(3x) =sin(30^@)#

This implies that

#3x=30^@ => x = 30^@/3 = 10^@#

Answer 2 · 2015-06-09T18:07:47

#10^@, 50^@#

#sin 3x = 1/2#
Trig table of special arcs gives
# 3x = 30^@#
#x = 10^@#

The trig unit circle gives another arc x that has the same sin value:
#3x = (180 - 30) = 150^@#
#x = 50^@#
Answers for (0, 360):
#10^@, 50^@#