Solving $5 \sin (2x)=1$ without a calculator? My friend asked me if I could help her with a few problems, and they are mostly quadratics of trig functions and easy manipulations. But I was surprised to find equations such as  $5 \sin (2x)=1$ and $10\sin(x-2)=-7$ because to me it seems you need to use inverse trig functions to solve this. The problem is, the problem set with inverse trig functions only starts later. Is there a way to solve these without explicitly using inverse trig functions? Thanks.
 A: From this answer: https://math.stackexchange.com/a/87768/26188 you get that $\sin(x)$ is never $\frac{1}{5}$ when $x$ is of the form $\frac{p\pi}{q}$ for any integers $p$ and $q$. So finding an exact solution to $5\sin(2x) = 1$ might be too much to ask of someone in pre-calculus. You have likewise for your second equation.
A: $\sin (2x) = 0.2$
Applying double angle formula, we have $2 \sin x \cos x = 0.2$
$\sin (x) \sqrt{1 – \sin^2{x}} = 0.1$
$\sin^2 (x) \left( 1 – \sin^2 x \right) = 0.01$
$\ -sin^4x + \sin^2x – 0.01 = 0$, which is now quadratic in $\sin^2x$.
Note 1. Normally this question should not be solved this way.
Note 2. Squaring will produce unnecessary roots.
Note 3. However, you still need the inverse of sine to find what x is.
Note 4. Course started on trigonometric identities?
Note 5. Double angle formula starts even later in the course.
A: $5 \sin (2x)=1 \Rightarrow \sin(2x)=\frac{1}{5}=\frac{opposite side}{hypoteneous}$
$\frac{CD}{DB}=\frac{AC}{AB} \Rightarrow \frac{1-a}{a}=\frac{5}{\sqrt(24)} \Rightarrow a=\frac{ \sqrt(24)}{5+ \sqrt(24)} \Rightarrow AD=\sqrt(24) \frac{\sqrt(1+(5+\sqrt(24))^2)}{5+\sqrt(24)} \Rightarrow \sin(x)=\frac{1}{\sqrt(1+(5+\sqrt24)^2)}$
Since $a$ is very small $\Rightarrow x$ is very small and for small angles $\sin x \approx x$
$\therefore x=\frac{1}{\sqrt(1+(5+\sqrt24)^2)} \approx 0.1$
