Let $G$ be a linear algebraic group over a field $k$ with separable closure $k^s$ and absolute Galois group $\Gamma\!_k$. Consider the Galois cohomology group $$ H^1(\Gamma\!_k,G(k^s)). $$ This group is known to be trivial in the following cases:
- $k$ arbitrary, $G = SL_n$ or $Sp_{2n}$ (Serre, Galois Cohomology, III.1).
- $k$ nonarchimedean local, $G$ semisimple and simply-connected (Bruhat-Tits).
Based on these examples, we should expect the following conjecture to hold.
Conjecture: Let $G$ be a simply-connected reductive algebraic group. Then $$ H^1(\Gamma\!_k,G(k^s)) = 1. $$
Is this conjecture known to be true? Is it known to be true with additional assumptions on $G$ or $k$?
As I mentioned in my comment, what you are asking is strongly related to Serre's Conjecture. In general though, this cohomology set is not trivial.
As an example, if $G=\operatorname{Spin}_n$ then $H^1(k,G)$ more or less classifies quadratic forms of dimension $n$ with trivial discriminant and trivial Clifford invariant. But not all such forms are hyperbolic for a general field $k$. For instance, if $k=k_0(x,y,z)$, then the $3$-Pfister form $q=\langle 1,x\rangle \langle 1,y\rangle \langle 1,z\rangle$ is non-split.
This example gives a pretty good idea of why Serre's conjecture restricts to fields of cohomological dimension at most 2: here I used the fact that I had 3 independant indeterminates to construct my quadratic form.