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I'm wondering if there is an analytical solution to the equation $ \cos x + a\sin x = d$. I only have a little progress so far. First of all, the equation soluble at least for cases $a=0$ and $a=1$. If $a$ is $0$, then it reduces to $\cos x=d$ so $x=\arccos d$. If $a$ is $1$, then $$\cos x+\sin x=d$$ $$(\cos x+\sin x)^2=d^2$$ $$\cos^2 x+\sin^2 x+2\sin x\cos x=d^2$$ $$1+\sin 2x=d^2$$ $$\sin 2x=d^2-1$$ $$2x=\arcsin(d^2-1)$$ $$x=\frac12\arcsin(d^2-1)$$ There is one condition I can place on the general form. The first derivative can be used to find that the range of $\cos x + a\sin x$ is $\left[-\sqrt{1+a^2},\sqrt{1+a^2}\right]$ so the equation can't be solved unless d is in that range. $$-\sqrt{1+a^2}\le d\le\sqrt{1+a^2}$$ $$|d|\le \sqrt{1+a^2}$$ $$d^2\le \ 1+a^2$$ $$d^2-a^2\le 1$$ which implies that the magnitude of $d$ isn't allowed to be much larger than the magnitude of $a$. Is there a way to solve this equation completely?

Gary
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MumboJumbo
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    You have the right idea. In general, an equation of the form $a \cos x + b \sin x =c$ can be solved by dividing throughout by $\sqrt{a^2+b^2}$ and let $\cos \alpha=\frac{a}{\sqrt{a^2+b^2}}$ to get $$\cos(x-\alpha)=\frac{c}{\sqrt{a^2+b^2}}.$$ – Anurag A Jan 25 '23 at 10:18

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