Maximum value of $5\sin x - 12 \cos x + 1$

Question

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 ?

Answer 1

Note that we have the maximum value for $x-67.4°=90°$ indeed for $x-67.4°=270°$ we have that $\sin=-1$ is minimum.

Answer 2

For any $k \in \Bbb Z$, $$\sin(x-67.4°) = 1 \iff x - 67.4° = 90° + k(360°) \iff x = 157.4° + k(360°). $$

Answer 3

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}