isolating T1

[ijd5000]

hi,
I can't seem to find a way to solve for T1 in this equation. (from a physics problem)

T₁sin(x₁)-gm/T₁cos(x₁)=-tan(x₂​)

 

[wjm11]

hi,
I can't seem to find a way to solve for T1 in this equation. (from a physics problem)

T₁sin(x₁)-gm/T₁cos(x₁)=-tan(x₂​)

Since you have a T1 in the denominator, multiply both sides of the equation by T1. You will end up with a quadratic equation.

Collect all terms on one side of the equation, with 0 on the other. Solve using the Quadratic Formula.

 

[ijd5000]

Since you have a T1 in the denominator, multiply both sides of the equation by T1. You will end up with a quadratic equation.

Collect all terms on one side of the equation, with 0 on the other. Solve using the Quadratic Formula.

didn't come up with a squared term so can't use quadratic . did i do something wrong?

 

[wjm11]

didn't come up with a squared term so can't use quadratic . did i do something wrong?

Perhaps you have not used grouping symbols appropriately, and I misunderstood your equation.

As written "T₁sin(x₁)-gm/T₁cos(x₁)=-tan(x₂​)" means: (T₁sin(x₁)) - (gm/T₁)(cos(x₁) = -tan(x₂​)

What exactly is your equation?

Regardless, I still don't see how you did not get a quadratic. Please show your work.

 

[HallsofIvy]

Example:
if ax - bc/(ay) = k

then a^2xy - bc = ayk

If you can't follow that, you're not ready for your posted problem.

Good example, because it also shows the importance of grouping. If this were (ax- bc)/ay= k then multiplying each side by ay would bi e ax- bc= ayk.