Sine and cosine inequalities

Geometry Level 3

{ sin x cos x 1 1 3 cos x 1 sin x 1. \large{\begin{cases} \dfrac{\sin x}{\cos x - 1}\geq 1 \\ \dfrac{3\cos x - 1}{\sin x}\geq 1. \end{cases}}

Find x ( 0 , 36 0 ) x \in (0^\circ, 360^\circ) in degrees satisfying the system of inequalities above.


The answer is 270.

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1 solution

Chew-Seong Cheong
Oct 22, 2018

Relevant wiki: Half Angle Tangent Substitution

sin x cos x 1 1 Let t = tan x 2 2 t 1 + t 2 1 t 2 1 + t 2 1 1 1 t 1 1 t < 0 . . . ( 1 ) \begin{aligned} \frac {\sin x}{\cos x -1} & \ge 1 & \small \color{#3D99F6} \text{Let }t = \tan \frac x2 \\ \frac {\frac {2t}{1+t^2}}{\frac {1-t^2}{1+t^2}-1} & \ge 1 \\ - \frac 1t & \ge 1 \\ \implies -1 & \le t < 0 & ...(1)\end{aligned}

Similarly,

3 cos x 1 sin x 1 3 3 t 2 1 t 2 2 t 1 2 4 t 2 2 t 1 2 t 2 + t 1 0 ( t + 1 ) ( 2 t 1 ) 0 t 1 t 1 2 . . . ( 2 ) \begin{aligned} \frac {3\cos x-1}{\sin x} & \ge 1 \\ \frac {3-3t^2-1-t^2}{2t} & \ge 1 \\ \frac {2-4t^2}{2t} & \ge 1 \\ 2t^2+t - 1 & \ge 0 \\ (t+1)(2t-1) & \ge 0 \\ \implies t \le -1 \cup t & \ge \frac 12 & ...(2) \end{aligned}

Comparing ( 1 ) (1) and ( 2 ) (2) , we note that there is only one solution t = tan x 2 = 1 x = 270 t= \tan \dfrac x2 = -1 \implies x = \boxed{270}^\circ ,

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