bamakhosimz

2022-09-11

Solve the Equation

$$\sqrt{3}tgx-\sqrt{3}ctgx=2$$

$$\sqrt{3}tgx-\sqrt{3}ctgx=2$$

Yareli Hendrix

Beginner2022-09-12Added 8 answers

$$\sqrt{3}tgx-\sqrt{3}ctgx=2$$

$$\sqrt{3}tgx-\frac{\sqrt{3}}{tgx}=2$$

tgx=t

$$\sqrt{3}t-\frac{\sqrt{3}}{t}=2$$

$$\sqrt{3}{t}^{2}-2t-\sqrt{3}=0$$

$$D=4+4\ast \sqrt{3}\ast (-\sqrt{3})=4+4\ast 3=16$$

$${t}_{1}=\frac{2+4}{2\sqrt{3}}=\frac{3}{\sqrt{3}}=\sqrt{3}$$

$${t}_{2}=\frac{2-4}{2\sqrt{3}}=-\frac{1}{\sqrt{3}}$$

$$tgx=\sqrt{3}$$

$$x=arctg(\sqrt{3})+\pi \ast n$$

$$x=\frac{\pi}{3}+\pi \ast n$$

$$tgx=-\frac{1}{\sqrt{3}}$$

$$x=arctg(-\frac{1}{\sqrt{3}})+\pi \ast k$$

$$x=-\frac{\pi}{6}+\pi \ast k$$

n and k $$\in $$ Z.

$$\sqrt{3}tgx-\frac{\sqrt{3}}{tgx}=2$$

tgx=t

$$\sqrt{3}t-\frac{\sqrt{3}}{t}=2$$

$$\sqrt{3}{t}^{2}-2t-\sqrt{3}=0$$

$$D=4+4\ast \sqrt{3}\ast (-\sqrt{3})=4+4\ast 3=16$$

$${t}_{1}=\frac{2+4}{2\sqrt{3}}=\frac{3}{\sqrt{3}}=\sqrt{3}$$

$${t}_{2}=\frac{2-4}{2\sqrt{3}}=-\frac{1}{\sqrt{3}}$$

$$tgx=\sqrt{3}$$

$$x=arctg(\sqrt{3})+\pi \ast n$$

$$x=\frac{\pi}{3}+\pi \ast n$$

$$tgx=-\frac{1}{\sqrt{3}}$$

$$x=arctg(-\frac{1}{\sqrt{3}})+\pi \ast k$$

$$x=-\frac{\pi}{6}+\pi \ast k$$

n and k $$\in $$ Z.

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