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OpenStudy (anonymous):
find all solutions 2sin^2x+3sinx+1=0
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OpenStudy (jdoe0001):
\(\bf 2sin^2(x)+3sin(x)+1=0\implies 2[{\color{brown}{ sin(x)}}]^2+3[{\color{brown}{ sin(x)}}]+1=0\) notice is just a quadratic equation.... thus factor it and solve for "sin(x)"
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