# Find coordinate points where the tangent line is horizontal for

Find coordinate points where the tangent line is horizontal for $f\left(x\right)=-\mathrm{sin}\left(8x\right)+6\mathrm{cos}\left(4x\right)-8x$
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Palandriy0

You have used the wrong double angle formula, it should be
$\mathrm{cos}8x=1-2{\mathrm{sin}}^{2}\left(4x\right)$
and corresponding quadratic equation would be $2{u}^{2}-3u-2=0$ and at last you would get $\mathrm{sin}4x=-\frac{1}{2}$

###### Not exactly what you’re looking for?
pacetfv
Let
$g\left(x\right)=f\left(\frac{x}{4}\right)=-\mathrm{sin}2x+6\mathrm{cos}x-2x$
Then
${g}^{\prime }\left(x\right)=-2\mathrm{cos}2x-6\mathrm{sin}x-2$
$=-2\left(1-2{\mathrm{sin}}^{2}x\right)-6\mathrm{sin}x-2$
$=2\left(2{\mathrm{sin}}^{2}x-3\mathrm{sin}x-2\right)$
$=2\left(2\mathrm{sin}x+1\right)\left(\mathrm{sin}x-2\right)$
Hence the critical points of g occur when $\mathrm{sin}x=-\frac{1}{2}$ (it is not possible for $\mathrm{sin}x=2$). There are four such $x\in \left(-\pi ,2\pi \right)$
$x\in \left\{-\frac{5\pi }{6},-\frac{\pi }{6},\frac{7\pi }{6},\frac{11\pi }{6}\right\}$
Consequently, the critical points of f occur when $\mathrm{sin}4x=-\frac{1}{2}$, and for $-\frac{\pi }{4}, this occurs at
$x\in \left\{-\frac{5\pi }{24},-\frac{\pi }{24},\frac{7\pi }{24},\frac{11\pi }{24}\right\}$