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Question #c2359

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Mar 5, 2015

Find the angles first (#3x#), then find #x#.

The cosine of some angle is to be #sqrt3/2#.

The angles with that cosine are all of the form: #pi/6+2pik# or #-pi/6+2pik# for integer #k#.

This tells us that #3x# must be #pi/6+2pik# or #-pi/6+2pik# for integer #k#.

Solve the two equations

#3x=pi/6+2pik# and #3x=-pi/6+2pik# for #x#.

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