# Find coordinate points where the tangent line is horizontal for $f(x) = –\sin(8x) + 6\cos(4x) – 8x$

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.

• $\cos8x=1-2\sin^24x$ – Nosrati Jul 3 '19 at 4:02
• I don't think. I've found $-2u^2+3u+2=0$ – Nosrati Jul 3 '19 at 4:04

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 $$\sin 4x = - \frac{1}{2}$$
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 $$x \in \left\{ -\frac{5\pi}{24}, - \frac{\pi}{24}, \frac{7\pi}{24}, \frac{11\pi}{24} \right\}.$$