Chapter 2 - Limits - 2.8 Intermediate Value Theorem - Exercises - Page 86: 8

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Let $f(x)=\sin nx-\cos x .$ Note that $f$ is continuous on $\left[0,\pi\right]$, with $f(0)=0-1=-1$ and $f(\pi)=0-(-1)=1 .$ Therefore, by the IVT, there is a $c \in\left[0,\pi\right]$ such that $f(c)=\sin nc-\cos c=0 .$ Thus $\sin nc-\cos c=0$ and hence the equation $\sin nx-\cos x=0$ has a solution $c$ in $\left[0,\pi\right]$.