Question

How would you solve something like `asint=bt` or `acost=bt` for `t`?

For example,
`8cosx=3x`
`5siny=5y`

Answer 1

Such equations generally have no algebraic solution, so you need to use a numerical method.

Explanation:

As Steve said, you would solve such equations using a numerical method such as Newton-Raphson:

Given a differentiable function `f(x)` for which we want to find a zero, start with a first approximation `a_0`, then repeatedly apply the formula:

`a_(i+1) = a_i - (f(a_i))/(f'(a_i))`

For example:

To find approximations to `x` satisfying `8cos(x) = 3x`, define:

`f(x) = 8cos(x)-3x`

Then:

`f'(x) = -8sin(x)-3`

Put the following formula into a spreadsheet:

`a_(i+1) = a_i + (8cos(x)-3x)/(8sin(x)+3)`

Then we find:

`a_0 = 0`

`a_1 = 2.bar(6)`

`a_2 ~~ 0.3965853162202127`

`a_3 ~~ 1.4128671965703734`

`a_4 ~~ 1.1394457744448268`

`a_5 ~~ 1.1322818941294521`

`a_6 ~~ 1.1322734714493981`

`a_7 ~~ 1.1322734714376355`

`a_8 ~~ 1.1322734714376355`

So it converges in a few steps to a good approximation of a (in fact the) solution.

If there is more than one solution, then you might try different values for `a_0`.