三角形ABC中,A>B且滿足方程8(sinx)^2+3sin2x-4=0,求C過程 謝謝~~
熱心網友
3sin2x-4(1-2(sinx)^2)=03sin2x-4cos2x=05(sin2xcos53-cos2xsin53)=05sin(2x-53)=0所以 2x-53=0 or 2x-53=180 x1=53/2 x2=233/2C=180-(53/2+233/2)= 37
三角形ABC中,A>B且滿足方程8(sinx)^2+3sin2x-4=0,求C過程 謝謝~~
3sin2x-4(1-2(sinx)^2)=03sin2x-4cos2x=05(sin2xcos53-cos2xsin53)=05sin(2x-53)=0所以 2x-53=0 or 2x-53=180 x1=53/2 x2=233/2C=180-(53/2+233/2)= 37