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

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

1 Answer
George C.
Apr 10, 2017

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#.

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