Solve for x given {0 < x < 2}? #sin(2x) + cos(x) = 0#

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Answer 1 · 2018-06-12T16:20:53

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

#sin2x+cosx=0--(1)#

#"for "{0 < x<2pi}#

#" now "sin2x=2sinxcosx#

#(1)rarr2sinxcosx+cosx=0#

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

#"either "cosx=0#

#=>x=pi/2, (3pi)/2#

#"or "2sinx+1=0#

#=>sinx=-1/2#

#:. x=(7pi)/6, (11pi)/6#

solution set

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