Solve 0.7 < \alpha < 0.8 for x in y = \cos 3x + 2

[Hootenanny]

The curve y = \cos 3x + 2 intersects the line y = 2x at point A, whose x co-ordinate is \alpha. Show that 0.7 &lt; \alpha &lt; 0.8.

So far I've got: Upon intersection 2x = \cos 3x + 2 \Rightarrow \cos 3x - 2x = - 2. This doesn't seem to help. I know we've done this type of thing ages ago, but I've since lost my notes and my minds gone blank. Any help would be appreciated.

 

[Galileo]

How about evaluating both functions at x=0.7 and x=0.8? See what you can do with that.

 

[Hootenanny]

\cos(3 \times 0.7) +2 = 1.495... , \cos(3 \times 0.8) +2 = 1.262....
2 \times 0.7 = 1.4, 2 \times 0.8 = 1.6.
All in radians. This doesn't seem to help??

 

[arildno]

Sure it does!
At x=0.7, we have the value as given by the straight line LOWER than that given by the cosine expression, whereas this is reversed at x=0.8
What does that tell you?

 

[Hootenanny]

Ahhh, ofcourse! Tha x - value must lie sumwhere between them values! I wan looking for an exact solution. Thank's foryou help guys!

 

[arildno]

You're welcome.
Most equations cannot be solved for an exact solution in a finite number of steps.
Approximative techniques abound, though.