Solving the trigonometric equation $2(1 + \sin \theta )\sin \theta =1$

Question

I usually use guessing to solve equations with trigonometric functions. Yesterday, I came across an equation that I could not really write it in a helpful form to guess. My question is, how can I solve equation like that without WolframAlpha? I really want to learn a method to solve trigonometric equations without guessing.

Some of my tires to write in a form that allows me to guess an answer. \begin{align} 2(1+\sin \theta )\sin \theta &= 1 \\ \sin \theta +\sin^2 \theta &= \frac{1}{2} \\ 2\sin \theta +(1-\cos 2\theta ) &= 1 \\ 2\sin \theta -\cos 2\theta &= 0 \\ 2\sin \theta -\cos^2 \theta +\sin^2 \theta &= 0 \\ \end{align}

Thank you.

Answer 1

Here you have a quadratic in $\sin\theta$ (second line)

$$\sin(\theta)+\sin^2(\theta) = \frac{1}{2} \iff \sin^2\theta + \sin \theta - 1/2 = 0$$

Set $x = \sin\theta$

$$x^2 + x - 1/2 = 0$$

$$x = \sin \theta = \dfrac{-1\pm\sqrt3}2$$

However, note that $\sqrt 3 > 1\implies -1-\sqrt3<-2\implies \dfrac{-1-\sqrt 3}2 < -1$

Since we need $\theta$ such that $ \; |\sin \theta| \leq 1$, we want only $\sin\theta = \dfrac{-1 + \sqrt 3}2$

Answer 2

So, $2\sin^2\theta+2\sin\theta-1=0$

Using the formula of Quadratic Eqaution,

$\sin\theta =\frac{-2\pm\sqrt{2^2-4(2)(-1)}}{2\cdot2}=\frac{-1\pm\sqrt3}2$

But as $\sqrt3>1\implies -1-\sqrt3<-2\implies \frac{-1-\sqrt3}2<-1$

For real $\theta,-1\le \sin\theta\le1$ $\implies \sin\theta=\frac{\sqrt3-1}2$