How do you solve 6x + 12 + 2cosx = 0?

Answer 1

6x +12 + 2cosx = 0
or
2cosx = -6x - 12
No straight-forward or easy way to get a reasonably accurate number.
Let's start by looking at the graphs of
y = 2cosx and y = -6x-12

http://www.wolframalpha.com/input/?i=plot+y+%3D+2cosx+,+y+%3D+-6x-12

looks like there is a solution near -2

how about Newton's Method:
dy/dx = 6 - 2sinx

newx = x - (6x+12+2cosx)/(6-2sinx)
where x is some starting value
begin with x = -2 , set your calculator to radians
newx ---- x
-1.893549... -1.894099..
-1.894099.. -1.8940998...
not bad after 2 steps.

x = appr -1.8941

verification:
http://www.wolframalpha.com/input/?i=sove+6x+%2B+12+%2B+2cosx+%3D+0

Answer 2

Newton's method converges rapidly. If you don't have calculus, then there are other methods, such a bisection.

f(x) = 6x + 12 + 2cosx

Start off with the interval (-3,-1) and find f at the midpoint: -2

f(-2) = 6(-2)+12+2cos(-2) = -0.832
So, f=0 occurs in (-2,-1).

f(-1.5) = 3.141
oops, too far, so f=0 in (-2,-1.5)

f(-1.75) = 1.14
f=0 in (-2,-1.75)

f(-1.875) = 0.151

Getting closer, but you can see that it will take a lot longer than Newton.