solve: cos3x.cos^3 x+sin 3x.sin^3 x=0.


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Question Number 30087 by rahul 19 last updated on 16/Feb/18

$solve :$ $\cos 3 x . {c o s}^{3} x + \sin 3 x . \sin^{3} x = 0 .$
	Answered by MJS last updated on 16/Feb/18

	$x = \frac{π}{4} + \frac{π \cdot z}{2}; z \in Z$ $\sin \frac{π}{4} = \cos \frac{π}{4} \land \sin \frac{3 π}{4} = - \cos \frac{3 π}{4}$ $you can easily see this looking$ $at the angles in a circle$
	Answered by MJS last updated on 16/Feb/18

	$. . . the hard way :$ $let me write c for \cos and s for \sin$ $c (α + β) = c (α) c (β) - s (α) s (β)$ $β = 2 α$ $c (3 α) = c (α) c (2 α) - s (α) s (2 α)$ $c (2 α) = {2 c}^{2} (α) - 1$ $s (2 α) = 2 s (α) c (α)$ $c (3 α) = {2 c}^{3} (α) - c (α) - 2 c (α) s^{2} (α)$ $s^{2} (α) = 1 - c^{2} (α)$ $c (3 α) = {4 c}^{3} (α) - 3 c (α)$ $s (α + β) = s (α) c (β) + c (α) s (β)$ $β = 2 α$ $s (3 α) = c (α) s (2 α) + c (2 α) s (α)$ $using c (2 α), s (2 α) as above$ $s (3 α) = ({4 c}^{2} (α) - 1) s (α)$ $c (3 α) c^{3} (α) + s (3 α) s^{3} (α) = 0$ ${4 c}^{4} (α) - {3 c}^{2} (α) + ({4 c}^{2} (α) - 1) s^{4} (α)$ $s^{4} (α) = {(1 - c^{2} (α))}^{2}$ ${8 c}^{6} (α) - {12 c}^{4} (α) + {6 c}^{2} (α) - 1 = 0$ $c (α) = \sqrt{t}$ ${8 t}^{3} - {12 t}^{2} + 6 t - 1 = 0$ $t^{3} - \frac{3}{2} t^{2} + \frac{3}{4} t - \frac{1}{8} = 0$ $here you need a good eye, the$ $left side is$ ${(t - \frac{1}{2})}^{3}$ $so t_{1} = t_{2} = t_{3} = \frac{1}{2}$ $\cos (α) = \frac{2 \sqrt{2}}{2}$ $α = \frac{π}{4} + \frac{π z}{2}; z \in Z$
	Commented by rahul 19 last updated on 16/Feb/18

	$thanks .$
	Answered by ajfour last updated on 16/Feb/18

	$(\cos 3 x) (4 \cos^{3} x) + (\sin 3 x) (4 \sin^{3} x) = 0$ $\cos 3 x (\cos 3 x + 3 \cos x) +$ $(\sin 3 x) (3 \sin x - \sin 3 x) = 0$ $2 \cos^{2} 3 x + 6 \cos x \cos 3 x$ $+ 6 \sin 3 x \sin x - 2 \sin^{2} 3 x = 0$ $1 + \cos 6 x + 3 (\cos 4 x + \cos 2 x)$ $+ 3 (\cos 2 x - \cos 4 x) - (1 - \cos 6 x) = 0$ $2 \cos 6 x + 6 \cos 2 x = 0$ $\Rightarrow 3 \cos 2 x + \cos 6 x = 0$ $o r 3 \cos 2 x + 4 \cos^{3} 2 x - 3 \cos 2 x = 0$ $\Rightarrow \cos^{3} 2 x = 0$ $o r \cos 2 x = 0$ $1 - 2 \sin^{2} x = 0$ $\sin^{2} x = \frac{1}{2}$ $x = n π \pm \frac{π}{4} .$
	Commented by rahul 19 last updated on 16/Feb/18

	$thank u sir .$
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