Find $\sin \theta$ and $\cos \theta$ given $\tan 2\theta$ Can you guys help with verifying my work for this problem. My answers don't match the given answers. 

Given $\tan 2\theta = -\dfrac{-24}{7}$, where $\theta$ is an acute angle, find $\sin \theta$ and $\cos \theta$

I used the identity, $\tan 2\theta = \dfrac{2\tan \theta}{1 - tan^2 \theta}$ to try and get an equation in $\tan \theta$.
$$
\begin{align}
-\dfrac{24}{7} &= \dfrac{2\tan \theta}{1 - \tan^2 \theta} \\
-24 + 24\tan^2 \theta &= 14 \tan \theta \\
24tan^2 \theta - 14\tan \theta - 24 &= 0 \\
12tan^2 \theta - 7\tan \theta - 12 &= 0 \\
\end{align}
$$
Solving this quadratic I got,
$$ \tan \theta = \dfrac{3}{2} \text{ or } \tan \theta = -\dfrac{3}{4}$$
$$\therefore \sin \theta = \pm \dfrac{3}{\sqrt{13}} \text{ and } \cos \theta = \pm \dfrac{2}{\sqrt{13}}$$
Or,
$$\therefore \sin \theta = \pm \dfrac{3}{5} \text{ and } \cos \theta = \mp \dfrac{4}{5}$$
The given answer is,

$$\sin \theta = \dfrac{4}{5} \text{ and } \cos \theta = \dfrac{3}{5}$$

I thought I needed to discard the negative solution assuming $\theta$ is acute. But they haven't indicated a quadrant. Do I assume the quadrant is I only? What am i missing? Thanks again for your help.
 A: At Chandru's request: 


*

*The quadratic $12z^2-7z-12$ factors as $(3z-4)(4z+3)$ so we should get $\tan\,\theta=4/3$ and $\tan\,\theta=-3/4$. 

*"Acute angle" means "angle between 0 and $\pi/2$" means 1st quadrant. 
A: Given $ \tan(2\theta)= -\frac{24}{7}$ From the relation between $\sin(\theta)$, $\cos(\theta)$ and $\tan(\theta)$, we get $$ \frac{\sin(2\theta)}{\cos(2\theta)}= -\frac{24}{7} \implies \sin(2\theta)= -\frac{24}{7} \cos(2\theta)$$and $$ \sin(2\theta)^2 + \cos(2\theta)^2=1$$ $$\cos(2\theta) = \pm \frac{7}{25} \implies 2 \cos^2(\theta)-1 = \pm \frac{7}{25}$$ Case 1: Rational number on the right is positive, $$\cos^2(\theta)=\frac{16}{25} \implies \cos(\theta) = \pm \frac{4}{5} $$ Solution to case 1: $$\cos(\theta)=\frac{4}{5}$$$$  \sin(\theta)=\frac{3}{5}.$$ Both sine and cosine functions are positive, for $\theta$ being acute.  Case 2:Rational number on the right is negative $$\cos^2(\theta)=\frac{9}{25} \implies \cos(\theta) = \pm \frac{3}{5} $$Solution to case 2: $$\cos(\theta)=\frac{3}{5} $$$$  \sin(\theta)=\frac{4}{5}.$$ Both sine and cosine functions are positive, for $\theta$ being acute. 
