How to get the exact value of $\sin(x)$ if $\sin(2x) = \frac{24}{25}$ How to get the exact value of $\sin(x)$ if $\sin(2x) = \frac{24}{25}$ ?
I checked various trigonometric identities, but I am unable to derive $\sin(x)$ based on the given information.
For instance:
$\sin(2x) = 2 \sin(x) \cos(x)$
 A: Since $\cos 2x=\pm\frac{7}{25}$, $\tan x=\frac{\sin 2x}{1+\cos 2x}\in\left\{\frac{24}{32},\,\frac{24}{18}\right\}=\left\{\frac{3}{4},\,\frac{4}{3}\right\}$ so $\sin x=\pm\frac{\tan x}{\sqrt{1+\tan^2 x}}\in\pm\left\{\frac{3}{5},\,\frac{4}{5}\right\}=\left\{-\frac{4}{5},\,-\frac{3}{5},\,\frac{3}{5},\,\frac{4}{5}\right\}$. 
A: $$|\sin(2x)|=|2\sin(x)\cos(x)|=|2\sin(x)\sqrt{1-\sin(x)^2}|$$
so
$$\left(\frac{24}{25}\right)^2=4\sin(x)^2(1-\sin(x)^2)$$
solve this quadratic for $\sin(x)^2$ and get $|\sin(x)|$. Both the positive and negative values will be possible (think of why).
A: $tanx=\frac{sin2x}{1+cos2x}$. You have $\sin2x$ and you can calculate $\cos2x$ with the Pythagorean theorem. Once you have $\ tanx$, you can use SOHCAHTOA. Don't forget the quadrants though
A: $$\cos ^2 {2x} = 1-\sin ^2 {2x} = \frac {49}{625}$$
$$ \cos 2x = \frac {7}{25}$$
$$ \sin ^ 2x = \frac {1-\cos 2x}{2} = 9/25 $$
$$ \sin x =  \frac {3}{5} $$ 
A: Refer to the diagram below.
$AD$ is the angle bisector of the right-angled triangle $\Delta ABC$.
Given $BC=24$ and $AC=25$.
Let $\angle DAB = x$.
From $\Delta ABC$ we see that $\sin(2x) = \dfrac{24}{25}$.
Now, by angle bisector theorem, $BD:DC = 7:25$.
Therefore, $BD = \dfrac{7}{7+25} \times 24 = \dfrac{21}4$.
Observing that $AB:BD = 4:3$, we see that $AD = \dfrac{35}4$
Therefore, $\sin x = \dfrac {BD} {DA} = \dfrac 3 5$.
The other possible $x$ would be $x+180^\circ$, since the period of sine is $360^\circ$, making $\sin x = -\dfrac35$.
In conclusion, $\sin x = \pm \dfrac 35$.

A: $$\sin(2x)=\frac{24}{25}$$
The identities you want are:
$$\sin(2x)=2\sin(x)\cos(x) \space [I]$$
and
$$\sin^2(x)+\cos^2(x)=1\space[II]$$
Square $[I]$ to get:
$$4\sin^2(x)\cos^2(x)=\frac{576}{625}\to\sin^2x(1-\sin^2x)=\frac{144}{625}$$
$$\to\sin^4(x)-\sin^2x+\frac{144}{625}=0$$
Let $\sin^2(x)=\sigma$. Your equation becomes:
$\sigma^2-\sigma+\frac{144}{625}=0$
Solve via Quadratic Formula:
$$\sigma=\frac{1\pm\sqrt{1-\frac{576}{625}}}{2}=\frac 12\pm\frac{\sqrt{\frac{49}{625}}}{2}=\frac{9}{25}, \frac{16}{25}.$$
Since $\sin^2(x)=\frac{9}{25}, \frac{16}{25}, \sin(x)=\pm \frac 35, \pm \frac 45$
Check these values with the function $f(S)=\sin(2\cdot\sin^{-1}(S))$, (where $S$ is the solutions i.e. $\pm\frac 35, \pm \frac 45$) to see the ones which provide the correct solution, and note the values which work are $\frac 35, \frac 45$.
We can see them graphed here:
