How do i calculate the following?

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Answer 1 · 2018-04-04T12:52:13

#alpha=15^@,75^@,25.3bar3^@,85.3bar3^@#

I am not sure of your notation here. The use of the comma to me, signifies:

#tan3alpha-1=0 \ \ \ \ [1]#

and:

#tan3alpha-1=3 \ \ \ \ \ [2]#

#[1]#

#tan3alpha-1=0#

#tan3alpha=1#

#3alpha=arctan(tan3alpha)=arctan(1)#

#3alpha=arctan(1)# #-> arctan(1)=45^@, 225^@#

These angles lie in the I and III quadrants.

So we have:

#3alpha=45=>alpha=45/3=15^@#

#3alpha=225=>alpha=225/3=75^@#

#[2]#

#tan3alpha-1=3#

#tan3alpha=4#

#3alpha=arctan(tan(3alpha))=arctan(4)#

#3alpha=arctan(4)# -> #arctan(4)~~76^@, 256^@#

#alpha=arctan(4)/3~~76/3=25.3bar3#

#alpha=arctan(4)/3~~256/3=85.3bar3#

We need:

#0<=alpha<=90^@#

#alpha=15^@,75^@,25.3bar3^@,85.3bar3^@#

Exact solutions:

#alpha=15^@,75^@,(arctan(4)/3)^@, (180+arctan(4)/3)^@#