<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] converting the given equation into $$\tan(x/2)$$ we get
$$8\,{\frac {1- \left( \tan \left( x/2 \right)  \right) ^{2}}{1+ \left( 
\tan \left( x/2 \right)  \right) ^{2}}}-12\,{\frac {\tan \left( x/2
 \right) }{1+ \left( \tan \left( x/2 \right)  \right) ^{2}}}-5
=0$$
setting $$\tan(x/2)=t$$ you have to solve this equation
$$-{\frac {13\,{t}^{2}+12\,t-3}{{t}^{2}+1}}=0$$ for $$t$$
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