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Problem: Consider the function $f(x) = –\sin(8x) + 6\cos(4x) – 8x$ where $–\pi/4 < x < \pi/2$. Find the exact $x$-coordinates of the points on the graph of f at which there is a horizontal tangent line.

This is my attempt.

Find derivative $$ f’(x) = -\cos(8x)\cdot8 – 6\sin(4x)\cdot4 – 8 $$ $$f’(x) = -8\cos(8x) – 24\sin(4x) – 8$$ $$ f’(x) = -8(\cos(8x) + 3\sin(4x))$$ $$f’(x) = 0 $$

$$0 = -8(\cos(8x) + 3\sin(4x) + 1)$$ Use double angle formula $1-\sin(4x)$ to replace $\cos(8x)$ $$0 = -8(1 – \sin^2(4x) + 3\sin(4x) + 1)$$ Let $u = \sin(4x)$ $$ 0 = 1 – u^2 + 3u + 1$$ $$ 0 = -u^2 + 3u + 2$$

Use quadratic formula to solve for $u$. $$\sin(x) = (3\pm \sqrt{17}) / 2$$ Answer can only be negative due to out of bounds so $\sin(x) = (3-\sqrt17) / 2$

Multiply equation by $4$ because $\cos(4x) $

$$\sin(4x) = [4\cdot(3-\sqrt{17})] / 2$$ I tried multiplying by $4$ but $\sin$ becomes out of bounds

So I just tried to get the inverse of $\sin(x)$ to find the reference angle. $$\sin^{-1}(\frac{3–\sqrt{17}}{2}) = -0.596\text{rad} = -34^\circ$$

$$x = (\pi + 0.596), (2\pi-0.596), (-0.596), (-\pi+0.596)$$

After graphing these tangent lines in a calculator, the points are close but not exact and thus are not horizontal. I've done other horizontal tangent line problems but this one, it seems I have to use the quadratic formula to find $\sin(x)$, rather than factoring.

Any advice on what I'm doing wrong would be greatly appreciated.

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    $\begingroup$ $\cos8x=1-2\sin^24x$ $\endgroup$ – Nosrati Jul 3 '19 at 4:02
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    $\begingroup$ I don't think. I've found $-2u^2+3u+2=0$ $\endgroup$ – Nosrati Jul 3 '19 at 4:04
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You have used the wrong double angle formula, it should be $$ \cos 8x = 1 - 2 \sin^{2} (4x) $$ and corresponding quadratic equation would be $ 2u^2 - 3u - 2= 0$ and at last you would get $\sin 4x = - \frac{1}{2}$

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Let $$g(x) = f(x/4) = -\sin 2x + 6 \cos x - 2x.$$ Then $$\begin{align*} g'(x) &= -2 \cos 2x - 6 \sin x - 2 \newline &= -2 (1 - 2 \sin^2 x) - 6 \sin x - 2 \newline &= 2 (2 \sin^2 x - 3 \sin x - 2) \newline &= 2 (2 \sin x + 1)(\sin x - 2). \end{align*}$$ Hence the critical points of $g$ occur when $\sin x = -1/2$ (it is not possible for $\sin x = 2$). There are four such $x \in (-\pi, 2\pi)$: $$x \in \left\{-\frac{5\pi}{6}, - \frac{\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}\right\}.$$ Consequently, the critical points of $f$ occur when $\sin 4x = -1/2$, and for $-\pi/4 < x < \pi/2$, this occurs at $$x \in \left\{ -\frac{5\pi}{24}, - \frac{\pi}{24}, \frac{7\pi}{24}, \frac{11\pi}{24} \right\}.$$

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