How do you find the exact value of #4cos^2theta+4sintheta-5=0# in the interval #0<=theta<360#?

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Answer 1 · 2016-12-17T14:27:26

#theta=30^o# or #150^o#

#4cos^2theta+4sintheta-5=0#

#hArr4(1-sin^2theta)+4sintheta-5=0#

or #-4sin^2theta+4sintheta-1=0#

or #4sin^2theta-4sintheta+1=0#

or #(2sintheta-1)^2=0#

i.e. #2sintheta-1=0# or #sintheta=1/2=sin30^o# or #sin150^o#

and #theta=30^o# or #150^o#