# Find $x$ for $3 \sec x \tan x - 4 \csc x \cot x = 0$

I am trying to solve this but get stuck, perhaps I am doing it wrong. Thanks for any help. $$3\sec x \tan x - 4 \csc x \cot x = 0$$ $$3 \cdot \frac{1}{\cos x} \cdot \frac{\sin x}{\cos x} - 4 \cdot \frac{1}{\sin x}\cdot \frac{\cos x}{\sin x} = 0$$ $$\frac{3\sin x}{\cos^{2}x} - \frac{4\cos x}{\sin ^{2}x} = 0$$ $$\frac{3\sin^{3} x - 4\cos^{3} x}{\cos^{2}x \sin^{2}x} = 0$$ $$\frac{3\sin^{3} x - 4\cos^{3} x}{\cos^{2}x (1 - \cos^{2}x)} = 0$$ $$3\sin^{3} x - 4\cos^{3} x= 0$$

• so $3 \sin^3 x = 4 \cos^3 x$ where neither one is allowed to be zero, then $\frac{\sin^3 x}{\cos^3 x} = \tan^3 x = \frac{4}{3}$ so .. Apr 5, 2023 at 3:16

Continuing your work: $$3\sin^3x-4\cos^3x=0$$ $$\implies\tan^3x=\frac{4}{3}$$ Assuming $$x\in\mathbb{R}$$, so that $$\tan x\in\mathbb{R}$$: $$\implies\tan x=\sqrt[3]{\frac{4}{3}}$$ $$\implies \boxed{x=\arctan\left(\sqrt[3]{\frac{4}{3}}\right)}+\pi n,n\in\mathbb{N}$$
• How does $sin - cos$ become $tan$? Apr 5, 2023 at 4:05
• Divide by $\cos^3x$ Apr 5, 2023 at 4:21