Conceptually, I find this confusing.
I would prefer a graphical explanation instead of a algebraic one.
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$\begingroup$ The x-intercept occurs when $y=0$, hence we need to solve for $x$, given $a\cos x - b= 0$. This means $a\cos x = b \iff \cos x = \frac ba$, and so $x = \cos^{-1} (\frac ba)$. $\endgroup$– amWhyCommented Oct 6, 2020 at 17:22
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$\begingroup$ The graph for $\cos x$ and $y$ is a straight line. $\endgroup$– cosmo5Commented Oct 6, 2020 at 17:27
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1 Answer
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The $x$-intercepts of $y=a\cos x -b$ are infinite $$x=\pm\arccos\left(\frac{b}{a}\right)+2k\pi;\;k\in\mathbb{Z}$$
Consider $y=4\cos x-3$. As $y=0$ we have $x=\pm\arccos\left(\frac{3}{4}\right)+k\pi\approx \pm 0.72+2k\pi$.
Some intercepts are $\ldots,-5.56, -0.72, 0.72, 5.56, 7.00,\ldots$.
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