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July 6, 2015

July 6, 2015

Posted by **sylvia ** on Sunday, November 27, 2011 at 4:17pm.

- precalulas -
**Reiny**, Sunday, November 27, 2011 at 5:23pmThere is no easy method to solve equations of this type.

One way is to graph the left side and the right side as individual functions and see where they intersect

e.g.

graph : y = 2x and y = cosx

you will see that they intersect appr. at x = 1/2

As a matter of fact that is how "Wolfram" shows the solution

http://www.wolframalpha.com/input/?i=2x%3Dcos%28x%29+

With a good calculator at hand, you can then pinpoint the answer more accurately

e.g.

x = .5, LS = 1, RS = .877 , RS < LS so lower the x

x = .4, LS = .8, RS = .921 RS > LS so raise the x

x = .45 , LS = .9, RS = .90044 , not bad , RS > LS , so raise the x a bit

x = .452, LS = .904 , RS = .8995 , RS < LS, lower the x

x = .4505, LS = .901 , RS = .9002

do you get the idea.

Wolfram had it at x = .450184

then LS = .900368

RS = .900367 , how is that for close