What is wrong with this approach to easy trig question?

Question

Solve this: $\dfrac{1-\cos x}{\sin x} = \dfrac{\sqrt{3}}{3} $

Of course this is true: $\dfrac{1-\cos x}{\sin x} = \tan\dfrac{x}{2} = \dfrac{\sqrt3}{3}$

But why is this approach invalid: $\dfrac{1-\cos x}{\sin x}=\dfrac{\sqrt3}{3} = \dfrac{\frac{1}{3}}{\frac{1}{\sqrt{3}}} $

So it follows $\sin x = \dfrac{1}{\sqrt{3}}$ and $\cos x = 1 - \dfrac{1}{3} = \dfrac{2}{3}$. Consequently, $\tan x = \dfrac{\sqrt{3}}{2}$

I'm embarrassed I didn't know how to explain why the approach above is wrong. Please advise.

Answer 1

$\displaystyle \frac{a}{b}=\frac{c}{d}$ does not imply that $a=c$ and $b=d$.

If $\sin x=\frac{1}{\sqrt{3}}$ and $\cos x=\frac{2}{3}$, then $\sin^2x+\cos^2x\ne1$.

Answer 2

$$1-cos x=1-(1-2sin^2\frac{x}{2})=2 sin^\frac{x}{2}$$

$$\frac{1- cos x}{sin x}=\frac {2 sin^2 \frac{x}{2}}{2 sin \frac{x}{2} cos \frac{x}{2}}=tan \frac{x}{2}=\frac{\sqrt 3}{3}= \frac {1}{\sqrt 3}= \frac{\frac{1}{2}}{\frac{\sqrt 3}{2}}=\tan \frac{\pi}{6}$$

⇒ $$\frac {x}{2}=\frac{\pi}{6}$$

⇒ $$x=\frac {\pi}{3}$$

is primitive root and general solution is $x=2k \pi+\frac{\pi}{3}$.