$\tan(x) = 3$. Find the value of $\sin(x)$ I’m trying to figure out the value for $\sin(x)$ when $\tan(x) = 3$.  The textbook's answer and my answer differ and I keep getting the same answer. Could someone please tell me what I'm doing wrong? 
1.) $\tan(x) = 3$, then $\frac{\sin(x)}{\cos(x)} = 3$. 
2.) Then $\cos(x) = \frac{1}{3}\sin(x)$
3.) $\sin^2 (x) + \left(\frac{1}{3}\sin(x)\right)^2 = 1$    //Pythagorean identity substitution.
4.) $\left(\frac{4}{3}\sin(x)\right)^2 = 1$   //Combining like terms
5.) $\frac{16}{9}\sin^2(x) = 1$   //Square the fraction so I can move it later.
6.) $\sin^2(x) = \frac{1}{\frac{16}{9}}$ //Divide both sides by $\frac{16}{9}$
7.) $\sin^2(x) = \frac{1}{1} * \frac{9}{16} = \frac{9}{16}$  //divide out the fractions
8.) $\sin(x) = \pm \frac{3}{4}$ //square root both sides.
So $\sin(x) = \pm \frac{3}{4}$ but the book says this is wrong. Any help is much appreciated.

If $\tan(x)=3$, then $\tan^2(x)=9$. This means that $\frac{\sin^2x}{1-\sin^2x}=9$. So, $\sin^2(x)=\frac9{10}$; in other words (at least if we're on the first quadrant), $\sin(x)=\frac3{\sqrt{10}}$.
Your error lies in item 4: $\sin^2(x)+\left(\frac1{3\sin(x)}\right)^2=\frac{10}9\sin^2(x)$.
