<s>[INST] Maximum value of $5\sin x - 12 \cos x + 1$ Given that $5\sin x - 12\cos x = 13\sin (x-67.4)$ 
Find the maximum value of $5\sin x - 12 \cos x + 1 $ and the corresponding value of x from 0 to 360. 
Maximum value = $13+1=14$ 
Corresponding value of $x$ 
$13\sin (x-67.4) + 1 = 14$ 
$\sin(x-67.4) = 1 $
$x = 157.4 , 337.4 $ 
I found the value of $x$ and there’s 2 values. However , the answer is only$157.4$ why is that the case ? 
 [/INST] Let $\cos\phi=\dfrac{5}{13}$, $\sin\phi=\dfrac{12}{13}$
\begin{eqnarray}
y&=&13\left(\frac{5}{13}\sin x-\frac{12}{13}\cos x\right)+1\\
&=&13(\sin(x)\cos\phi-\cos(x)\sin\phi)+1\\
&=&13\sin(x-\phi)+1
\end{eqnarray}
which has a maximum value of $14$ when $x-\phi=\frac{\pi}{2}$.
So
\begin{eqnarray}
x&=&\frac{\pi}{2}+\phi\\
&=&\frac{\pi}{2}+\arctan\left(\frac{12}{5}\right)\\
&\approx&90^\circ+67.38^\circ\\
&=&157.38^\circ
\end{eqnarray}
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