<s>[INST] $\cos x − 2\sin x = 1$. How to solve using $R\cos(x+a)$? I have this question: $\cos x − 2\sin x = 1$. You have to solve for $x$, the answer being 4.069. I've tried answering but I just keep getting 2.214. This is my method:
$\cos x − 2\sin x = 1$
$R\cos(x+a)= 1$
$R = \sqrt{1^2 + (-2)^2}$
$a = \arctan(-2/1)$
therefore $\sqrt5 \cos(x - 1.107) = 1$
$\cos(x - 1.107) = \sqrt5/5$
$\arccos(\sqrt5/5) = x - 1.707$
$1.707 = x - 1.707$
therefore $x = 2.214$
Can anyone give me a hint where I'm going wrong? Thank you
 [/INST] $$\begin{align}
\cos x-2\sin x&=R\cos(x+\alpha)\\[1ex]
&=R\cos x\cos\alpha-R\sin x\sin\alpha
\end{align}$$
$$\implies\begin{cases}R\cos\alpha=1\\R\sin\alpha=2\end{cases}$$
Dividing the expressions in the second equation by those in the first gives
$$\frac{R\sin\alpha}{R\cos\alpha}=\tan\alpha=2\implies \alpha=\tan^{-1}(2)+n\pi$$
(where $n\in\mathbb Z$)
Taking the squares and summing them gives
$$(R\cos\alpha)^2+(R\sin\alpha)^2=1^2+2^2\implies R^2=5\implies R=\pm\sqrt5$$
Take $R=\sqrt5$ and $n=0$, so that
$$\cos x-2\sin x=\sqrt 5\cos(x+\tan^{-1}(2))=1$$
Can you take it from here?
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