Second-degree homogeneous equation

Statement

Solve the second-degree homogeneous equation: $\sqrt{3}\sin^2 x - (1+\sqrt{3})\sin x\cos x + \cos^2 x = 0$

Solution

Case $\cos x = 0$: substituting $x=\pi/2$ gives $\sqrt{3}\cdot 1 - 0 + 0 = \sqrt{3}\ne 0$. Not a solution.

Divide by $\cos^2 x$: $\sqrt{3}\tan^2 x - (1+\sqrt{3})\tan x + 1 = 0.$ With $t=\tan x$ we solve $\sqrt{3}\,t^2 - (1+\sqrt{3})t + 1 = 0$. The discriminant is $\Delta = (1+\sqrt{3})^2 - 4\sqrt{3} = 4 - 2\sqrt{3} = (\sqrt{3}-1)^2,$ so $t = \frac{(1+\sqrt{3}) \pm (\sqrt{3}-1)}{2\sqrt{3}} \;\Rightarrow\; t = 1 \ \vee\ t = \frac{1}{\sqrt{3}}.$

Back to $x$:

• $\tan x = 1$: $\ x = \dfrac{\pi}{4} + k\pi$;
• $\tan x = \dfrac{1}{\sqrt{3}}$: $\ x = \dfrac{\pi}{6} + k\pi$.

$\boxed{\,x = \frac{\pi}{4} + k\pi \ \vee\ x = \frac{\pi}{6} + k\pi,\quad k\in\mathbb{Z}.}$

Links

Topics: Trigonometric equations
Concepts: Homogeneous equation
Functions: Cosine · Sine · Tangent
Skills: Reasoning by cases · Solving equations · Using formulae
Exercise type: Solving an equation