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Question: How to solve an equation without RootOf

Question:How to solve an equation without RootOf

love-algebra 25 Maple 2016
solve rootof
November 10 2020
1

Hello
I solved an equation using Maple . but in answer there is RootOf statement that I can't change to a simple statement without RootOf.

restart

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((Lr2*(vo+vs)*sqrt(Lr1*Lr2)+vo*Lr1*sqrt(Lr1*Lr2))*sin(t*sqrt((Lr1+Lr2)/(Lr1*Lr2*cr)))*sqrt(cr*(Lr1+Lr2))+vs*t*Lr1*(Lr1+Lr2))/((Lr1+Lr2)^2*Lr1)

((Lr2*(vo+vs)*(Lr1*Lr2)^(1/2)+vo*Lr1*(Lr1*Lr2)^(1/2))*sin(t*((Lr1+Lr2)/(Lr1*Lr2*cr))^(1/2))*(cr*(Lr1+Lr2))^(1/2)+vs*t*Lr1*(Lr1+Lr2))/((Lr1+Lr2)^2*Lr1)``

 (1)

eq := ((Lr2*(vo+vs)*sqrt(Lr1*Lr2)+vo*Lr1*sqrt(Lr1*Lr2))*sin(t*sqrt((Lr1+Lr2)/(Lr1*Lr2*cr)))*sqrt(cr*(Lr1+Lr2))+vs*t*Lr1*(Lr1+Lr2))/((Lr1+Lr2)^2*Lr1)

((Lr2*(vo+vs)*(Lr1*Lr2)^(1/2)+vo*Lr1*(Lr1*Lr2)^(1/2))*sin(t*((Lr1+Lr2)/(Lr1*Lr2*cr))^(1/2))*(cr*(Lr1+Lr2))^(1/2)+vs*t*Lr1*(Lr1+Lr2))/((Lr1+Lr2)^2*Lr1)

 (2)

solve(eq, t)

-(Lr1*Lr2)^(1/2)*(Lr1*cr+Lr2*cr)^(1/2)*sin(RootOf(Lr1*sin(_Z)*(Lr1*cr+Lr2*cr)^(1/2)*((Lr1+Lr2)/(Lr1*Lr2*cr))^(1/2)*(Lr1*Lr2)^(1/2)*vo+Lr2*sin(_Z)*(Lr1*cr+Lr2*cr)^(1/2)*((Lr1+Lr2)/(Lr1*Lr2*cr))^(1/2)*(Lr1*Lr2)^(1/2)*vo+Lr2*sin(_Z)*(Lr1*cr+Lr2*cr)^(1/2)*((Lr1+Lr2)/(Lr1*Lr2*cr))^(1/2)*(Lr1*Lr2)^(1/2)*vs+_Z*Lr1^2*vs+_Z*Lr1*vs*Lr2))*(Lr1*vo+Lr2*vo+Lr2*vs)/(Lr1*(Lr1+Lr2)*vs)

 (3)

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Download spr.mw

How can I simplify to an answer without RootOf?

Thank u
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