$$x^2-4x-\cos x=-8$$ We want the number of solutions.
Now I tried taking $\cos x$ as constant but by formula for root we get a trig equation which can't be solved. Any help? Thanks!
2026-04-12 20:51:44.1776027104
How many solutions to $x^2-4x-\cos x=-8$?
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$$x^2-4x+8=\cos x$$ $$L.H.S=(x-2)^2+4\ge4$$ $$R.H.S\le1$$ Therefore, the number of solutions is zero.