<s>[INST] Solving trig equation with two ways gives two different answers It is given that $x$ satisfies
$$ 3\sin x + 4\cos x = 5 $$
We have to find the value(s) of $\tan (x/2) $. My approach is this:
$$\begin{align} &3\sin x + 4\cos x = 5 \\
&\implies \dfrac 3 5\sin x + \dfrac 4 5 \cos x = 1 \\
&\implies \cos ( x - \arccos(\dfrac 4 5) ) = 1 \\
&\implies x = 2n\pi + \arccos(\dfrac 4 5) \\
&\implies x/2 = n\pi + 0.5\arctan(\dfrac 3 4)\\
&\implies \tan(x/2) = + \tan(0.5 \arctan(\dfrac 3 4)) = y \;\; \text{ (say)}
\end{align}$$
Now, $$ \dfrac{2y}{1-y^2} = \tan (\arctan(\dfrac 3 4 ) ) =  \dfrac 3 4 $$
Which gives me $ y = 3, -\dfrac 1 3 $
However, the solution given does something like this:
$$\begin{align} &3\sin x + 4\cos x = 5 \\
&\implies \dfrac 3 5 \dfrac{ 2t } { 1 + t^2 } + \dfrac 4 5 \dfrac{ 1 - t^2 } { 1 + t^2 } = 1 \;\;\;\;(t = \tan (x/2))\\
&\implies (3t-1)^2 = 0 \implies t = \dfrac 1 3
\end{align}$$
Which one is correct and why?
 [/INST] There is a sign error. The solution of $ \dfrac{2y}{1-y^2} =  \dfrac 3 4 $ is $ y = -3, \dfrac 1 3 $.
</s>