How do you solve #sec ((5 theta)/4) =2# from 0 to 2pi?

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Answer 1 · 2016-10-11T16:43:56

#S={48^@,240^@,336^@}#

#sec ((5theta)/4)=2#

#1/cos ((5theta)/4)=2#

#1=2 cos((5theta)/4)#

#1/2 = cos((5theta)/4)#

#cos^-1 (1/2) =((5theta)/4)#

#+- 60^@ + 360n^@=(5theta)/4#

#4/5(+- 60^@ + 360n^@)=theta#

#+-48^@+288n^@=theta#

#n=0, theta=+- 48^@#

#n=1, theta = 336^@, 240^@#

#S={48^@,240^@,336^@}#