Library / Harvard-MIT Math Tournament /17 of 83

Harvard-MIT Math Tournament

Algebra Difficulty 4.3 AIME Prove it United States

Problem:
Find all $x$ between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ such that $1-\sin^{4} x-\cos^{2} x=\frac{1}{16}$ .

Solution

Solution:
$1-\sin^{4} x-\cos^{2} x = \frac{1}{16} \Rightarrow (16-16 \cos^{2} x) - \sin^{4} x - 1 = 0 \Rightarrow 16 \sin^{4} x - 16 \sin^{2} x + 1 = 0$ .

Use the quadratic formula in $\sin x$ to obtain $\sin^{2} x = \frac{1}{2} \pm \frac{\sqrt{3}}{4}$ .

Since $\cos 2x = 1 - 2 \sin^{2} x = \pm \frac{\sqrt{3}}{2}$ , we get $x = \pm \frac{\pi}{12}, \pm \frac{5\pi}{12}$ .

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.