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