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

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    $\begingroup$ 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 .. $\endgroup$
    – Will Jagy
    Apr 5, 2023 at 3:16

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Continuing your work: $$3\sin^3x-4\cos^3x=0$$ $$\implies\tan^3x=\frac{4}{3}$$ Assuming $x\in\mathbb{R}$, so that $\tan x\in\mathbb{R}$: $$\implies\tan x=\sqrt[3]{\frac{4}{3}}$$ $$\implies \boxed{x=\arctan\left(\sqrt[3]{\frac{4}{3}}\right)}+\pi n,n\in\mathbb{N}$$

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  • $\begingroup$ How does $sin - cos$ become $tan$? $\endgroup$
    – Rolomoto
    Apr 5, 2023 at 4:05
  • $\begingroup$ Divide by $\cos^3x$ $\endgroup$
    – DanDan面
    Apr 5, 2023 at 4:21

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