Solving $\int^b_0x=\int^b_0\cos x$ 
$R$ is the region below the curve $y=x$ and above the $x$-axis from $x=0$ to $x=b$, where $b$ is a positive constant. $S$ is the region below the curve $y=\cos x$ and above the $x-axis$ from $x=0$ to $x=b$. For what value of $b$ is the area of $R$ equal to the area of $S$?   

This is the question I was given. I took this and set up an equation: $$\int^b_0x=\int^b_0\cos x$$I then took the integrals getting: $$\frac{b^2}{2}=\sin b$$ I solved for $b$ and got $$b=\sqrt{2\sin b}$$ The question given to me had multiple choice answers, so I just plugged them all in until I got the right answer - $1.404$.  The other multiple choice options were: $$.739$$$$.877$$$$.986$$$$4.712$$
However, I am left with a question:Is there any way to solve this without being given multiple choice options?
 A: If you had a calculator to check you answer, you had one to solve the problem with some numerical method. The most basic and simple one would be Fixed Point Iteration. You'd get a good enough answer in 4 steps. Particularly, this is easy if your calculator allows you to write sqrt(2*sin(ANS)) and press enter repeatedly to keep iterating.
See below the results of this iteration:
$$ \begin{matrix}
b & \sqrt{2\sin(b)}\\
1         &1,297282533\\
1,297282533 &1,387679869\\
1,387679869 &1,402341597\\
1,402341597 &1,404168809\\
1,404168809 &1,404385791\\
1,404385791 &1,404411400\\
1,404411400 &1,404414420\\
1,404414420 &1,404414776\\
1,404414776 &1,404414818\\
1,404414818 &1,404414823 \end{matrix}$$
A: As MathIsFun commented, graphing or by inspection you notice that the solution is close to $\frac \pi 2$. Building the simplest Taylor series, you would get
$$\frac{b^2}{2}-\sin b=\left(\frac{\pi ^2}{8}-1\right)+\frac{1}{2} \pi  \left(b-\frac{\pi
   }{2}\right)+O\left(\left(b-\frac{\pi
   }{2}\right)^2\right)$$
giving as a very first approximation
$$b_{(1)}=\frac{8+\pi ^2}{4 \pi }\approx 1.42202$$ Using one more term for the series expansion
$$\frac{b^2}{2}-\sin b=\left(\frac{\pi ^2}{8}-1\right)+\frac{1}{2} \pi  \left(b-\frac{\pi
   }{2}\right)+\left(b-\frac{\pi }{2}\right)^2+O\left(\left(b-\frac{\pi
   }{2}\right)^4\right)$$ giving as a second approximation
$$b_{(2)}=\frac{1}{4} \left(\pi +\sqrt{16-\pi ^2}\right)\approx 1.40439$$ For sure, you could use Newton method which, starting with $b_0=\frac \pi 2$ would generate the following iterates
$$\left(
\begin{array}{cc}
n & b_n \\
 0 & 1.570796327 \\
 1 & 1.422017936 \\
 2 & 1.404656638 \\
 3 & 1.404414871 \\
 4 & 1.404414824
\end{array}
\right)$$
To get "nice looking" approximations, instead of Taylor series, you could use $[1,n]$ Padé approximants which would write
$$\frac{b^2}{2}-\sin b\sim\frac {\left(\frac {\pi^2}8-1\right)+a_1^{(n)}\left(b-\frac{\pi
   }{2}\right)} {1+\sum_{p=1}^n c_p^{(n)}\left(b-\frac{\pi
   }{2}\right)^p}$$ giving as estimates
$$b_{(n)}=\frac \pi 2- \frac{\frac {\pi^2}8-1 }{a_1^{(n)}}$$  For example, using $n=2$ would give 
$$b_{(2)}=\frac{64+32 \pi ^2-\pi ^4}{64 \pi }\approx 1.40444$$
