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Answer to Question #10998 in Trigonometry for kanika arora

Question #10998
solve for x: 2cos^2x+3sinx=0
Expert's answer
cos^2(x)=1-sin^2(x)
2*cos^2(x)+3sin(x)=0
2(1-sin^2(x))+3sin(x)=0
2*sin^2(x)-3sin(x)-2=0
sin(x)=t
then
2*t^2-3t-2=0
t=2 or t=-0.5
|sin(x)|<=1 so t=2 haveno
solution
sin(x)=-0.5 then x=-pi/3+2*pi*n, x=7/3*pi+2*pi*n where n is integer

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