What is the general solution of #2tan^2 3x-sec3x=1#?

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Answer 1 · 2017-11-22T16:38:38

#x=120^@xxn+-16.06^@# or #x=120^@xxn+60^@#, where #n# is an integer.

#2tan^2 3x-sec3x=1# can be written as

#2(sec^2 3x-1)-sec3x=1#

or #2sec^2 3x-sec3x-3=0#

or #(2sec3x-3)(sec3x+1)=0#

i.e. #sec3x=3/2=sec48.19^@# i.e. #3x=360^@xxn+-48.19^@#

i.e. #x=120^@xxn+-16.06^@#

or #sec3x=-1=sec180^@# i.e. #3x=360^@xxn+180^@#

i.e. #x=120^@xxn+60^@#, where #n# is an integer.