All the roots of $5\cos x - \sin x = 4$ in the interval $0^{\circ} \leq x \leq 360^{\circ}$?

Question

This is a problem that I stumbled upon in one of my books.

Representing $5\cos x - \sin x$ in the form $R\cos(x + \alpha)$ (as demanded by the question):

$ \rightarrow R = \sqrt{5^2 + ({-}1)^2} = \sqrt{26}\\ \rightarrow R\cos x \cos \alpha - R\sin x \sin \alpha = 5\cos x - \sin x \\ \rightarrow ➊\hspace{0.25cm}5 = \sqrt{26}\cos \alpha \\ \rightarrow ➋\hspace{0.25cm}{-}1 = \sqrt{26}\sin \alpha \\ $

So here are my question(s): $\\$

• Why is only ➊ working out?

• When I find one solution in the interval, which is 27°, why is another solution like $(360 - x)$ not working out (in this case 333°)?

• I see 310.4° as a solution to this equation in my book. How do you get to 310.4° from 27° (which is the solution I got)? Which $\cos$ identity am I missing? And surprisingly, my book doesn't include 27° as a solution!

Answer 1

\begin{eqnarray} 5\cos x - \sin x &=& 4\\ \frac{5}{\sqrt{26}}\cos x-\frac{1}{\sqrt{26}}\sin x&=&\frac{4}{\sqrt{26}}\\ \cos\alpha\cos x-\sin\alpha\sin x&=&\frac{4}{\sqrt{26}}\\ \end{eqnarray}

Giving

$$\cos(x+\alpha)=\frac{4}{\sqrt{26}}$$

So either

$$ x+\alpha=\arccos\left(\frac{4}{\sqrt{26}}\right)=38.33^\circ $$

or

$$ x+\alpha=360^\circ-\arccos\left(\frac{4}{\sqrt{26}}\right)=321.67^\circ $$

with

$$\alpha=\arccos\left(\frac{5}{\sqrt{26}}\right)=11.31$$

So you get that either $x=27.02^\circ$ or $x=310.36^\circ$.

Answer 2

It seems that you know why $R=\sqrt{26}$,
Hence $5\cos x - \sin x$ can be expressed as $\sqrt{26}(\frac{5}{\sqrt{26}}\cos x - \frac{1}{\sqrt{26}}\sin x)$ $$\text{Let, } \cos\alpha = \frac{5}{\sqrt{26}} \text{ and } \sin\alpha = \frac{1}{\sqrt{26}}$$ Hence, $5\cos x - \sin x$ can now be expressed as $\sqrt{26}\cos (x + \alpha)$.

To solve your initial equation, we can now write it as, $$\sqrt{26}\cos (x + \alpha) =4$$ $$\cos (x + \alpha) = \frac{4}{\sqrt{26}}$$ $$x +\alpha = \cos^{-1} \frac{4}{\sqrt{26}}$$ If $\cos\alpha = \frac{5}{\sqrt{26}}$, then $\alpha = \cos^{-1} \frac{5}{\sqrt{26}}$, $$x = \cos^{-1} \frac{4}{\sqrt{26}} - \alpha \rightarrow x = \cos^{-1} \frac{4}{\sqrt{26}} - \cos^{-1} \frac{5}{\sqrt{26}}$$ $$\therefore x = 27^o \text{ or } 321.40^o$$

Answer 3

This is just my way of saying what everyone else has already said.

\begin{align} R &=\sqrt{1^2+5^2}=\sqrt{26} \\ \cos \alpha &= \dfrac{5}{\sqrt{26}} \\ \sin \alpha &= \dfrac{1}{\sqrt{26}} \\ \alpha &\approx 11.31^\circ \\ \hline 5\cos x - 1 \sin x &= 4 \\ \cos x \cos \alpha - \sin x \sin \alpha &= \dfrac{4}{\sqrt{26}} \\ \cos(x + \alpha) &= \dfrac{4}{\sqrt{26}} \\ x + 11.31^\circ &\approx \pm 38.33^\circ + n 360^\circ \\ \hline x &\approx 38.33^\circ - 11.31^\circ = 27.02^\circ \\ x &\approx -38.33^\circ - 11.31^\circ + 360^\circ = 310.36^\circ \end{align}