How do you solve #tan^2x+3secx+3=0# and find all solutions in the interval #0<=x<360#?

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Answer 1 · 2016-09-23T15:36:47

Solution is #x=120^o# or #x=180^o# or #x=240^o#

#tan^2x+3secx+3=0#

#hArrsec^2x-1+3secx+3=0#

or #sec^2x+3secx+2=0#

or #sec^2x+secx+2secx+2=0#

or #secx(secx+1)+2(secx+1)=0#

or #(secx+2)(secx+1)=0#

i.e. #secx=-2# or #secx=-1#

If #secx=-2#, in the range #0^o,=x,360^o#,

#x=120^o# or #x=240^o# and

if #secx=-1#, #x=180^o#

Hence solution is #x=120^o# or #x=180^o# or #x=240^o#