How do you solve #\sin ^ { 2} x + 3\sin x + 2= 0#?

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Answer 1 · 2016-11-12T17:04:06

#x=2npi+(3pi)/2#

Factorizing #sin^2x+3sinx+2=0#

#sin^2x+2sinx+sinx+2=0#

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

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

Now as #sinx+2!=0#, we have #sinx+1=0#

or #sinx=-1=sin((3pi)/2)#

Hence, #x=2npi+(3pi)/2#