cosx=sin3x find x with solution pllllllz


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Question Number 60536 by hovea cw last updated on 21/May/19

$cosx = \sin 3 x$ $find x with solution$ $pllllllz$
	Answered by mr W last updated on 21/May/19

	$\cos x = \sin (\frac{π}{2} - x) = \sin 3 x$ $\Rightarrow 3 x = \frac{π}{2} - x + 2 n π$ $\Rightarrow x = \frac{n π}{2} + \frac{π}{8}, n ϵ Z$ $o r$ $\Rightarrow 3 x = (2 n + 1) π - (\frac{π}{2} - x)$ $\Rightarrow x = n π + \frac{π}{4}, n ϵ Z$
	Commented by hovea cw last updated on 21/May/19

	$ok sir thanks$
	Answered by MJS last updated on 21/May/19

	$\cos x = \sin 3 x$ $x = \arctan t$ $\frac{1}{\sqrt{t^{2} + 1}} = \frac{t (3 - t^{2})}{\sqrt{{(t^{2} + 1)}^{3}}}$ $t^{2} + 1 = t (3 - t^{2})$ $t^{3} + t^{2} - 3 t + 1 = 0$ $(t - 1) (t^{2} + 2 t - 1) = 0$ $t_{1} = - 1 - \sqrt{2}; t_{2} = - 1 + \sqrt{2}; t_{3} = 1$ $x_{1} = \arctan (- 1 - \sqrt{2}) = - \frac{3 π}{8} \pm π n$ $x_{2} = \arctan (- 1 + \sqrt{2}) = \frac{π}{8} \pm π n$ $x_{3} = \arctan 1 = \frac{π}{4} \pm π n$ $[n \in N]$
	Commented by hovea cw last updated on 21/May/19

	$how did u get arctant = x$ $skr$
	Commented by MJS last updated on 21/May/19

	$I substituted it to get a solveable equation$ $in t$
	Commented by MJS last updated on 22/May/19

	$\sin 3 x = - \sin x (1 - {4 \cos}^{2} x)$ $\sin \arctan t = \frac{t}{\sqrt{t^{2} + 1}}$ $\cos \arctan t = \frac{1}{\sqrt{t^{2} + 1}}$ $\sin 3 x = - \frac{t}{\sqrt{t^{2} + 1}} (1 - \frac{4}{t^{2} + 1}) = \frac{t (3 - t^{2})}{\sqrt{{(t^{2} + 1)}^{3}}}$
	Answered by malwaan last updated on 22/May/19

	$c o s x = c o s (\frac{π}{2} - 3 x)$ $x = \pm (\frac{π}{2} - 3 x) + 2 n π$ $∴ x = \frac{π}{8} + \frac{n π}{2}; n \in Z$ $o r x = \frac{π}{4} - n π = \frac{π}{4} + n π; n \in Z$
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