Question
Solve the following equations:
$\tan\text{x}+\tan2\text{x}+\tan3\text{x}=0$
$\tan\text{x}+\tan2\text{x}+\tan3\text{x}=0$
✓
Answer
$\tan\text{x}+\tan2\text{x}+\frac{(\tan\text{x}+\tan2\text{x})}{1-\tan\text{x}.\tan2\text{x}}=0$
$[\tan\text{x}+\tan2\text{x}]\Big[1+\frac{1}{1-\tan\text{x}.\tan2\text{x}}\Big]=0$
$\tan\text{x}+\tan2\text{x}(2-\tan\text{x}.\tan2\text{x})=0$
$\tan\text{x}=\tan(-2\text{x})$ or $\tan\text{x}.\tan2\text{x}=0$
$\text{x}=\text{n}\pi-2\text{x}$ or $\tan\text{x}.\frac{2\tan\text{x}}{1-\tan^{2}\text{x}}=2$
$3\text{x}=\text{n}\pi$ or $\frac{2\tan^{2}\text{x}}{1-\tan^{2}\text{x}}=2$
$3\text{x}=\text{n}\pi$ or $2\tan^{2}\text{x}=2-2\tan^{2}\text{x}$
$3\text{x}=\text{n}\pi$ or $4\tan^{2}\text{x}=2$
$\text{x}=\frac{\text{n}\pi}{3}$ or $\tan^{2}\text{x}=\frac{1}{2}$
$\text{x}=\frac{\text{n}\pi}{3}$ or $\text{x}=\text{m}\pi\pm\tan^{-1}(\frac{1}{\sqrt{2}}),\text{n,m}\in\text{Z}$
$[\tan\text{x}+\tan2\text{x}]\Big[1+\frac{1}{1-\tan\text{x}.\tan2\text{x}}\Big]=0$
$\tan\text{x}+\tan2\text{x}(2-\tan\text{x}.\tan2\text{x})=0$
$\tan\text{x}=\tan(-2\text{x})$ or $\tan\text{x}.\tan2\text{x}=0$
$\text{x}=\text{n}\pi-2\text{x}$ or $\tan\text{x}.\frac{2\tan\text{x}}{1-\tan^{2}\text{x}}=2$
$3\text{x}=\text{n}\pi$ or $\frac{2\tan^{2}\text{x}}{1-\tan^{2}\text{x}}=2$
$3\text{x}=\text{n}\pi$ or $2\tan^{2}\text{x}=2-2\tan^{2}\text{x}$
$3\text{x}=\text{n}\pi$ or $4\tan^{2}\text{x}=2$
$\text{x}=\frac{\text{n}\pi}{3}$ or $\tan^{2}\text{x}=\frac{1}{2}$
$\text{x}=\frac{\text{n}\pi}{3}$ or $\text{x}=\text{m}\pi\pm\tan^{-1}(\frac{1}{\sqrt{2}}),\text{n,m}\in\text{Z}$
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