$\sin 2\theta +\sin \theta =1$ I tried Wolframalpha to solve this equation. The solution is $\theta\approx 0.355$. Since once a wise guy at MSE told me not to trust machines, I would like to know what methods can be used to solve this equation. I would appreciate for a brief explanation. 
By the way $0\leq\theta\leq\pi/4$.
My second question is for any $x>0$ and $x\in\mathbb{R}$, Does $$x\sin 2\theta +\sin \theta =1$$ have a real valued solution for $0\leq\theta\leq\pi/4$?

Well, note that if we put $t=\tan\frac\theta 2,$ then we have $$\begin{align}\sin\theta &= \sin\left(2\cdot\frac\theta2\right)\\ &= 2\sin\frac\theta2\cos\frac\theta2\\ &= 2\tan\frac\theta2\cos^2\frac\theta2\\ &= \cfrac{2t}{\sec^2\frac\theta2}\\ &= \cfrac{2t}{\tan^2\frac\theta2+1}\\ &= \frac{2t}{t^2+1},\end{align}$$ and in a similar fashion, we can show that $$\cos\theta=\frac{1-t^2}{t^2+1}.$$
Now, use the double-angle formula:
$$x\sin 2\theta+\sin\theta=1\\2x\sin\theta\cos\theta+\sin\theta=1\\2x\cdot\frac{2t}{t^2+1}\cdot\frac{1-t^2}{t^2+1}+\frac{2t}{t^2+1}=1\\\frac{4xt-4xt^3}{(t^2+1)^2}+\frac{2t}{t^2+1}=1\\0=1-\frac{4xt-4xt^3}{(t^2+1)^2}-\frac{2t}{t^2+1}\\0=1+\frac{4xt^3-4xt}{(t^2+1)^2}=\frac{-2t}{t^2+1}\\0=\frac{(t^2+1)^2}{(t^2+1)^2}+\frac{4xt^3-4xt}{(t^2+1)^2}+\frac{-2t(t^2+1)}{(t^2+1)^2}\\0=\frac{t^4+2t^2+1}{(t^2+1)^2}+\frac{4xt^3-4xt}{(t^2+1)^2}+\frac{-2t^3-2t}{(t^2+1)^2}\\0=\frac{t^4+(4x-2)t^3+2t^2-(4x+2)t+1}{(t^2+1)^2}$$
Since $t^2+1$ is nonzero for all real $t,$ then
$$0=t^4+(4x-2)t^3+2t^2-(4x+2)t+1.$$ Note that the coefficients of the above polynomial sum to $0,$ so $t=1$ is a root of the polynomial, and so $t-1$ is a factor. In particular, $$0=(t-1)\bigl(t^3+(4x-1)t^2+(4x+1)t-1\bigr),$$ so $$t=1\qquad\text{or}\qquad t^3+(4x-1)t^2+(4x+1)t-1=0.$$ The cubic equation can be solved using the general formula for roots of a cubic. (Warning: The roots thus found will probably be ugly-looking even if they happen to be rational numbers, and may look non-real even when they are actually real.)
Once you've found all your values of $t=\tan\frac\theta2,$ the values of $\theta$ will be those values of $2\arctan(t)$ lying in the appropriate interval, if there are any such values.
