已知cos(4分之拍πX)-5分之3,求1-tanX分之sin2X-2sin平方X的值
举一反三
- 求微分方程[img=634x60]17da653955cf9e7.png[/img]的特解。 （ ） A: sin(2*x)/3 - cos(x) - cos(x)/3 B: sin(2*x)/3 - cos(x) - sin(x)/3 C: cos(2*x)/3 - cos(x) - sin(x)/3 D: sin(2*x)/3 - sin(x) - sin(x)/3
- (-24分之5)x(-18分之15)x(-1又3分之2)计算
- （1）7X=5分之3（2）12分之5x=8分之3（3）X÷9分之4=12分之7（4）3分之2X÷4分之1=9分之8
- $\int {{1 \over {3 + 5\cos x}}} dx = \left( {} \right)$ A: ${1 \over 4}\ln \left| {{{2\cos x + \sin x} \over {2\cos x - \sin x}}} \right| + C$ B: ${1 \over 4}\ln \left| {{{2\cos {x \over 2} + \sin {x \over 2}} \over {2\cos {x \over 2} - \sin {x \over 2}}}} \right| + C$ C: $\ln \left| {{{\cos {x \over 2} + \sin {x \over 2}} \over {\cos {x \over 2} - \sin {x \over 2}}}} \right| + C$ D: $\ln \left| {{{\cos x + \sin x} \over {\cos x - \sin x}}} \right| + C$
- 求微分方程[img=143x21]17da5f14490e50e.png[/img]的通解,实验命令为(). A: dsolve(D2y-2*Dy+5*y=sin(2*x),x)ans =exp(x)*sin(2*x)*C2+exp(x)*cos(2*x)*C1+1/17*sin(2*x)+4/17*cos(2*x) B: dsolve('D2y-2*Dy+5*y=sin(2*x)','x')ans =cos(2*x)*(sin(4*x)/17 - cos(4*x)/68 + 1/4) - sin(2*x)*(cos(4*x)/17 + sin(4*x)/68) + C1*cos(2*x)*exp(x) - C2*sin(2*x)*exp(x) C: dsolve(D2y-2*Dy+5*y=sin(2*x),'x','y')ans =exp(x)*sin(2*x)*C2+exp(x)*cos(2*x)*C1+1/17*sin(2*x)+4/17*cos(2*x)