Sin2x+cosx=0 solve the equation on the interval 0,2pi?

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Answer 1 · 2018-05-14T04:39:59

#x=pi/2, (7pi)/6, (3pi)/2, (11pi)/6#

Recall the identity #sin2x=2sinxcosx#. Rewrite the equation with this identity:

#2sinxcosx+cosx=0#

And note that we can factor out #cosx:#

#cosx(2sinx+1)=0#

We now solve the equations #cosx=0, 2sinx+1=0:#

#cosx=0 -> x=pi/2, (3pi)/2# solve this on the interval #[0, 2pi).#

#2sinx+1=0#

#2sinx=-1#

#sinx=-1/2#

#x=(7pi)/6, (11pi)/6# solve this on #[0, 2pi).#