<s>[INST] $\sin 2\theta +\sin \theta =1$ 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 $0\leq\theta\leq\pi/4$?
 [/INST] $$\begin{align}
x \sin 2\theta + \sin\theta = 1 \quad &\implies \quad 2 x \sin\theta \cos\theta = 1 - \sin\theta \\
&\implies \quad 4 x^2 \sin^2\theta \left(1-\sin^2\theta\right)=\left(1-\sin\theta\right)^2 \\
&\implies \quad \left( 1 - s \right) \left( 4 x^2 s^3 + 4 x^2 s^2 + s - 1 \right) = 0 \\
\end{align}$$
where $s := \sin\theta$. The first factor's root, $s = 1$, corresponds to $\theta = \pi/2 + 2 n \pi$, which is outside the specified domain of interest. Roots via the second factor
$$p(s) := 4 x^2 s^3 + 4 x^2 s^2 + s - 1 = 0$$
can be found via the messy cubic formula. Notice that, by the Descartes Rule of Signs, the polynomial $p$ has exactly one positive root (for any real $x$). Notice also that 
$$p(\sin 0) = p(0) = -1 \qquad \text{and} \qquad p\left(\sin\frac{\pi}{4}\right) = p\left(\frac{\sqrt{2}}{2}\right) = x^2 \left( 2 + \sqrt{2} \right) - \frac{1}{2}\left( 2 - \sqrt{2} \right)$$
The polynomial's single positive root lies between $\sin 0$ and $\sin\frac{\pi}{4}$ (inclusive) when, and only when, the signs at the endpoints disagree or the value at the larger endpoint vanishes; thus, that latter value must be non-negative,
$$x^2 \left( 2 + \sqrt{2} \right) \geq \frac{1}{2} \left( 2 - \sqrt{2} \right)$$
so that
$$x^2 \geq \frac{2 - \sqrt{2}}{2\left( 2 + \sqrt{2} \right)}=\frac{2-\sqrt{2}}{2\left(2 + \sqrt{2}\right)}\frac{2-\sqrt{2}}{2-\sqrt{2}} = \frac{\left( 2 - \sqrt{2}\right)^2}{4}$$
whereupon, the condition
$$|x| \geq \frac{2 - \sqrt{2}}{2}$$
is necessary and sufficient for the equation $x \sin 2\theta + \sin\theta = 1$ to have a solution in the interval $\left[0, \frac{\pi}{4}\right]$.
</s>