If $\theta$ is an acute angle, and $\sin 2\theta=a$, then $\sin\theta+\cos\theta$ equals
$\textbf{(A) }\sqrt{a+1}\qquad \textbf{(B) }(\sqrt{2}-1)a+1\qquad \textbf{(C) }\sqrt{a+1}-\sqrt{a^2-a}\qquad\\ \textbf{(D) }\sqrt{a+1}+\sqrt{a^2-a}\qquad  \textbf{(E) }\sqrt{a+1}+a^2-a$
Let $x = \sin\theta+\cos\theta$, so we want to find $x$. First, square the expression to get $x^2 = \sin^2 \theta + 2\sin\theta\cos\theta + \cos^2 \theta$. Recall that $\sin^2 \theta + \cos^2 \theta = 1$ and $\sin 2\theta = 2\sin\theta\cos\theta$. Plugging these in, the equation simplifies to $x^2 = 1 + \sin 2\theta$. Given that $\sin 2\theta=a$, the equation becomes $x^2=1+a$. Take the square root of both sides to get $x = \sqrt{1+a}$.
Hence, the answer is $\boxed{\textbf{(A) }\sqrt{a+1}}$ ~[jiang147369](https://artofproblemsolving.comhttps://artofproblemsolving.com/wiki/index.php/User:Jiang147369)