Solving for $a$ in this equation $\sin^a(a) = b$ $$\sin^{a}a=b$$
Do you know how to solve for $a$ here algebraicaly? This is a little bit too abstract for my understanding. Thank you very much.
In the same spirit as amcalde's answer, and staying in the real domain which implies $a\in[2k \pi,(2k+1)\pi]$ (this as been pointed out by Quang Hoang in a comment)  and $b\in[0,1]$.

Limiting the analysis to the case of $k=0$,  considering  $f(x)= \sin^x(x) $, you could easily notice that, starting at $x=0$ $(f(0)=1)$, it decreases and reaches a minimum value close to $x_1\approx 0.398576$ $(f_1=f(x_1)\approx 0.68575)$, starts increasing up to $x_2=\frac \pi 2$ $(f(x_2)=1)$ and then decreases to $0$ for $x_3=\pi$ $(f(x_3)=0)$.

So, depending on the value of $b$, the equation $$\sin^x(x)=b$$ can have


two solutions if $b=1$ (namely $0$ and $x_2$)
three  solutions if $f_1&lt;b&lt;1$
two solutions if $b=f_1$
one solution if $b&lt;f_1$


As said, only numerical methods (such as Newton) will solve the problem but the question is to have a reasonable starting value.

Just as amcalde answered, look at the plot, make a guess $x_0$ for the root(s) you want and start Newton iterations $$x_{n+1}=x_n-\frac{\sin ^{x_n}(x_n)-b}{\sin ^{x_n}(x_n) (x_n \cot (x_n)+\log (\sin (x_n)))}$$ to the required accuracy.

For illustration purposes, consider $b=0.8$. The plot shows roots close to $0.1$, $0.9$, and $2.0$. 

Applying the method for the first root, the iterates will be $$x_1=0.0944109$$ $$x_2=0.0945459$$ $$x_3=0.0945460$$ which is the solution for six significant figures.

Applying the method for the second root, the iterates will be $$x_1=0.892865$$ $$x_2=0.892846$$ 

Applying the method for the third root, the iterates will be $$x_1=2.03211$$ $$x_2=2.03102$$ 

As you see, the solutions are obtained quite fast.

Edit

After Quang Hoang's good point, I had a look at the case $k&gt;0$ which is quite interesting. For any value of $b\in[0,1]$, it seems that there are always two roots on each side of $(4k+1)\frac \pi 2$. The curves are sharper and sharper picks (they look like gaussian curves in spite of some skewness that numerical calculations can easily show - at mid height, corresponding to $b=0.5$, the distance between the two roots is more or less $2 \sqrt{\frac{\log (2)}{1+k\pi}}$ ) and the problem becomes simpler than for the case $k=0$ since, for $x\in[2k \pi,(2k+1)\pi]$, we can approximate very accurately $$\sin^x(x)\approx \exp \left(-(1+k\pi) \left(x-(4k+1)\frac{\pi}{2}  \right)^2\right)$$from which quite good estimates can be generated for each of the two roots. $$x_{\pm}=(4k+1)\frac{\pi}{2}\pm \sqrt{\frac{-\log(b)}{1+k\pi}}$$ For example, still using $b=0.8$ and $k=1$ the above approximation gives as estimates $x_-=7.62186$, $x_+=8.08610$ while the solutions are $7.61304$ and $8.08781$. To some extent, the formula also applies for the estimates of the largest solution when $k=0$ (for the case looked at, it gives as estimates $1.09842$ and $2.04318$).

On the other side, the cases of $k&lt;0$ seem to be nightmares (it does not seem that solutions exist - According to Quang Hoang's comment, this is wrong : there are effectively two solutions if $b&gt;1$).

Edit

For the case of $k=0$ and $0.68575&lt;b&lt;1$, the initial (very crude) estimates can be obtained 


For the first root, $$x=x_1 \left(1-\sqrt{\frac{b-f_1}{1-f_1}}\right)$$ This was obtained after fitting the function by a quadratic polynomial going through points $(0,1)$, $(x_1,f_1)$ with an horizontal tangent at  $(x_1,f_1)$.
For the second root,$$x=\frac{\pi}{2}- \sqrt{-\log(b)}$$
For the third root,$$x=\frac{\pi}{2}+ \sqrt{-\log(b)}$$


For the case $b=0.8$, this wil generate the estimates $0.16$, $1.10$ and $2.04$ which seem to be quite reasonable; starting withe these values, Newton method converges in very  few iterations. 

For the case where $k=0$ and $0&lt;b&lt;f_1$ the estimate of the solution is
$$x=\pi -(\pi -x_4) \sqrt{\frac{b}{f_1}}$$  This has been obtained after approximating the function by a quadratic polynomial going through points $( x_4,f_1)$, $( \pi ,0)$ with horizontal tangents at the end point ($x_4 \approx 2.14647$ corresponds to the largest solution of $\sin^x(x)=f_1$).

Let us try Newton method for $b=0.1$. The successive iterates are then $$x_0=2.76158$$ $$x_1=2.69242$$ $$x_2=2.70099$$ $$x_3=2.70112$$ 

Edit

The case to also consider is if $b$ is a very small number. The approximation given from the tails of the approximating gaussian curves is not very good. In such a case, what is prosed is to use as starting values


for the smallest root,  $x=2k\pi+b^{\frac 1{2k\pi}}$
for the largest  root,  $x=(2k+1)\pi-b^{\frac 1{2k\pi}}$


For illustration purposes, consider $k=1$ and $b=10^{-4}$. The estimates are then $6.51406$ and $9.19390$ while the solutions are $6.39888$ and $9.20013$.

Edit

When we generate an estimate $x_0$, it must be such that $x_0\in[2k \pi,(2k+1)\pi]$. With the gaussian approximation, the value of the function is not zero at the bounds. So, with a minor sacrifice for the value of the maximum of the curve, I suggest to use instead 
$$\sin^x(x)\approx \exp \left(-(1+k\pi) \left(x-(4k+1)\frac{\pi}{2}  \right)^2\right)-\exp \left({-\frac{1}{4} \pi ^2 (1+k\pi )}\right)$$ which will give as estimates $$x_\pm=  \left(2 k+\frac{1}{2}\right)\pi\pm\sqrt{-\frac{\log \left(b+e^{-\frac{1}{4} \pi ^2
   (1+k\pi)}\right)}{1+k\pi  }}$$ which also works fine for small values of $b$.

Finally, in order to introduce some skewness in the gaussian, the function is defined independently for left and right sides according to $$\sin^x(x)\approx \exp \left(-\alpha \left(x-(4k+1)\frac{\pi}{2}  \right)^2\right)-\exp \left({-\frac{\alpha}{4} \pi ^2 }\right)$$ using $\alpha=\frac 34+k\pi$ for the left side and  $\alpha=\frac 54+k\pi$ for the right side of the curve. 

For illustration purposes, consider again $k=1$ and $b=10^{-4}$. The estimates are then $6.35931$ and $9.19390$ while the solutions are $6.39888$ and $9.28798$.
You have to use power reducing trig identities. Wikipedia has a list of them: https://en.wikipedia.org/wiki/List_of_trigonometric_identities. Go to the power reducing section and use either the one for odd sine powers or the one for even sine powers. The formula is rather messy, which is why it isn't typically a problem discussed much. Even after reducing the power, you'll probably have to use a fair amount of algebra, to massage things into a nice format.