$\sin 2\theta +\sin \theta =1$ 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 $0\leq\theta\leq\pi/4$?
 A: Here’s a hint: start by using the double-angle formula $$\sin 2\theta = 2\sin \theta \cos \theta,$$ then use the identity
$$\sin^2 \theta + \cos^2 \theta = 1.$$
A: 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.
A: $$\begin{align}
x \sin 2\theta + \sin\theta = 1 \quad &\implies \quad 2 x \sin\theta \cos\theta = 1 - \sin\theta \\
&\implies \quad 4 x^2 \sin^2\theta \left(1-\sin^2\theta\right)=\left(1-\sin\theta\right)^2 \\
&\implies \quad \left( 1 - s \right) \left( 4 x^2 s^3 + 4 x^2 s^2 + s - 1 \right) = 0 \\
\end{align}$$
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 $\left[0, \frac{\pi}{4}\right]$.
A: Let $f(\theta)=x\sin2\theta+\sin\theta-1$ , then if $\theta=0$ , $f(\theta)=-1 $ and if $\theta=\frac{\pi}{4}$ ,$f(\theta)=x+\frac{\sqrt{2}}{2}-1$ . Therefore if $x> 1-\frac{\sqrt2}{2}$ , by the Bolzano's theorem , exists a $\theta_{0}\in[0,\frac{\pi}{4}]$ such that $f(\theta_{0})=0$.
