How to solve for $x$ in exact terms where $2x^2\sin(x^2) = 3\cos(x^2)$ I want to know if it is possible to solve for $x$ without approximating. The equation is:
$$2x^2\sin(x^2) = 3\cos(x^2)$$
There are multiple solutions, but the answer I am looking for is about $0.9441$. Does anyone know if this is even possible? Thank you in advance :)
 A: There is not a closed form solution. 
The best we can do is to show that a solution exists and determine it by numerical methods.
For the first part let consider $y=x^2\ge0$ then
$$2y\sin y = 3\cos y \iff f(y)=\frac23y\tan y=1$$
from wich we can see that form IVT exactly one solution exists in the interval $y\in\left(0,\frac \pi 2\right)$ and exactly one solution for any other interval $y\in\left(\frac \pi 2+k\pi,\frac 32\pi +k\pi\right)$ with integer $k\ge 1$.
Note also that for $y$ large
$$\frac23y\tan y=1 \iff \frac23\tan y=\frac1y \to 0 $$
therefore solutions for $y$ approximate the values $\frac \pi 2 +k\pi$ for $k$ large.
A: (Edit: This might be viewed as the numerical technique in addition to user's theoretical justification)
There is no algebraic solution to this equation.
That is because you can not isolate $x$.
However you can numerically approximate a solution.
For example with Newton's method.
For that observe:
$f(x)=2x^2\sin(x^2)-3\cos(x^2)$
Now calculate $f'(x)$ and guess a start value for your approximation. Since you know the value already, you might take $x_0=1$.
Then iterate the formula:
$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$
So with $x_0=1$ you have $x_1=1-\frac{f(1)}{f'(1)}$ and do that over and over again until you reach a satisfying approximation.
A: As said, no analytical solution and numerical methods should be used.
However, you could have a good approximation of the first zero of function
$$f(y)=2y\sin (y)- 3\cos (y)$$ building its simplest Padé approximant "close" to the root $y=a$. This would write
$$f(y)\sim \frac{f(a)+ \left(f'(a)-\frac{f(a) f''(a)}{2 f'(a)}\right) (y-a)}   {1-\frac{ f''(a)}{2 f'(a)}(y-a) }$$ and setting the numerator equal to zero would then give
$$y=a+\frac{2 f(a) f'(a)}{f(a) f''(a)-2 f'(a)^2}$$
Using $a=\frac \pi 3$ would give
$$y=\frac \pi 3+\frac{270 \sqrt{3}-8 \pi  \left(18+\sqrt{3} \pi \right)}{1539+20 \pi  \left(3 \sqrt{3}+\pi \right)}\approx 0.988303$$ while the exact solution would be $\approx 0.988241$.
