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Question Number 72248 by Omer Alattas last updated on 26/Oct/19

Commented by Omer Alattas last updated on 26/Oct/19

$s o l v t o x$
	Answered by MJS last updated on 26/Oct/19

	$\tan^{2} x - 2 \sin 2 x = 0$ $x = \arctan t$ $t^{2} - 2 \frac{2 t}{t^{2} + 1} = 0$ $\frac{t^{2} (t^{2} + 1) - 4 t}{t^{2} + 1} = 0$ $t (t^{3} + t - 4) = 0$ $t_{1} = 0$ $t_{2} \approx 1.37880$ $t_{3, 4} \approx - . 689398 \pm 1 . 55750 i$ $x_{1} = n π$ $x_{2} \approx . 943311 + n π$
	Commented by Omer Alattas last updated on 27/Oct/19

	$t h a n k y o u p r o$
	Commented by Omer Alattas last updated on 26/Oct/19

	$h o w i c a n w r i t e t h e a n c c e r i n r a d p l e a s$
	Commented by MJS last updated on 27/Oct/19

	$x \in [0; 2 π]$ $x_{1} leads to 0; π; 2 π$ $x_{2} leads to . 943311; 4.08490$
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