<s>[INST] Solve equation $8\cos x - 6\sin x = 5$ where $0 \le x \le 360$ Solve equation $8\cos x - 6\sin x = 5$ where $0 \le x \le 360$.
I am not asked to use any form, so I am going to use $k\cos(x-\alpha)$.
$8\cos x - 6\sin x = k\cos(x-\alpha)$
$$=k(\cos x\cos\alpha + \sin x\sin\alpha)$$
$$=k\cos\alpha\cos x + k\sin\alpha\sin x$$
equating coefficients:
$k\cos\alpha = 8$
$k\sin\alpha = 6$, or is it $k\sin\alpha = -6$? I find this really confusing
$k = \sqrt{8^2 + 6^2} = 10$
$\alpha$ is in the 1st quadrant where both sin and cos are positive.
$\alpha = \arctan\frac{6}8 = 36.8^{\circ}$
$\therefore10\cos(x - 36.8) = 5$
I have a maximum and minimum value of 10
From here I do not know how to finish solving for x.
 [/INST] So, $\cos(x-\alpha)=\dfrac5{10}=\cos60^\circ$
$\implies x-\alpha=360^\circ n\pm60^\circ$ where $n$ is any integer
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