# 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

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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