How many solutions does this have?

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how many solutions does this equation have?

cos(3*x)-x=0

I only found one which was at x= about 1

Im just wondering if there are any others that i may have missed.
 
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set cos(3x)=x and graph the functions on the same plot. the number of intersections is the number of solutions.

that's the geometric way.

algebraically... you can replace cos(3x) by as many terms as you need from it's power series. so a second order approximation would just be 1-x^2/2! this is now a second order equation and has two solutions. since you found one of them to be real then the other one is real also. so I would expect at least 2 solutions. keep using better approximations for cos(3x) to see if you can find other solutions on the real line.
 
Well weird thing is that on my graphing calculator...i put that equation in and i just got a long straight line that crosses the x-axis around 1. But using Newton- Raphson method, i got an additional 3 points (-.93, -.84, .34) but i can't get the like .98 value that i see on my calculator.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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