How do you find the exact solutions of the equation #sin2x+cosx=0# in the interval #[0,2pi)#?

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Answer 1 · 2017-08-31T04:43:21

#x in{pi/2,(7pi)/6,(3pi)/2,(11pi)/6}#

We are seeking solution in interval #[0,2pi)#

As #sin2x+cosx=0#. we have

#2sinxcosx+cosx=0#

or #cosx(2sinx+1)=0#

i.e. either #cosx=0# meaning #x=pi/2,(3pi)/2#

or #2sinx+1=0# i.e. #sinx=-1/2# meaning #x=(7pi)/6,(11pi)/6#

Hence #x in{pi/2,(7pi)/6,(3pi)/2,(11pi)/6}#