$-1 = 4\cos(t^2) - \sin(t^2)\cos(t^2)$ $$-1 = 4\cos(t^2) - \sin(t^2)\cos(t^2)$$
I was solving a problem and I arrived to this. I am wondering if there is a way to solve for $t\in (1,2)$ without using a computer.
 A: It's not pretty.
You can try to divide both sides by $\cos^2 t^2$, and you get:
$$-\sec^2t^2=4\sec t^2-\tan t^2$$
Rewrite $\sec^2t^2$ as $\tan^2t^2+1$ and isolate $4\sec t^2$, you get:
$$-\tan^2t^2-1+\tan t^2=4\sec t^2$$
Square both sides and you get:
$$\tan ^4\left(t^2\right)-2 \tan ^3\left(t^2\right)+3 \tan ^2\left(t^2\right)-2 \tan \left(t^2\right)+1=16\sec^2 t^2\\
\tan ^4\left(t^2\right)-2 \tan ^3\left(t^2\right)-13 \tan ^2\left(t^2\right)-2 \tan \left(t^2\right)-15=0$$
You can then try solve the quartic equation in $\tan t^2$. 

Although, you can use numerical methods to solve this:
Using NR you get this recurrence relation for $f(\text{t})\text{:=}4 \cos \left(t^2\right)-\sin \left(t^2\right) \cos \left(t^2\right)+1$:
$$t_{n+1}=t_n-\frac{f(t_n)}{f'(t_n)}\iff \frac{1-\left(\sin \left(t_n^2\right)-4\right) \cos \left(t_n^2\right)}{2 t_n \left(4 \sin \left(t_n^2\right)+\cos \left(2 t_n^2\right)\right)}+t_n$$
After sufficient iterations you get solutions as:
$$\text{for}\,\, t_0=0.8\Rightarrow t=1.3799511968833622\\
\text{for}\,\, t_0=0.6\Rightarrow t=2.1237221862028157\\
\text{for}\,\, t_0=0.5\Rightarrow t=2.861372155620343\\
\vdots$$
You can find other $t$ in $[1,2]$, but I believe only one exists.
