8sin-3-x-pi-6-cos-3x-x-

Question Number 130819 by EDWIN88 last updated on 29/Jan/21

8 \sin^{3} (x + \frac{π}{6}) = \cos (3 x)

x = ?

Answered by mr W last updated on 29/Jan/21

l e t u = x + \frac{π}{6}

\Rightarrow 3 x = 3 u - \frac{π}{2}

\cos (3 x) = \cos (3 u - \frac{π}{2}) = \sin (3 u)

= 3 \sin u - 4 \sin^{3} u

8 \sin^{3} u = 3 \sin u - 4 \sin^{3} u

(4 \sin^{2} u - 1) \sin u = 0

\Rightarrow \sin u = 0 \Rightarrow u = k π \Rightarrow x = k π - \frac{π}{6}

\Rightarrow 4 \sin^{2} u - 1 = 0 \Rightarrow \sin u = \pm \frac{1}{2}

\Rightarrow u = 2 k π + \frac{π}{2} \pm \frac{π}{3} \Rightarrow x = 2 k π + \frac{π}{3} \pm \frac{π}{3}

\Rightarrow u = 2 k π - \frac{π}{2} \pm \frac{π}{3} \Rightarrow x = 2 k π - \frac{2 π}{3} \pm \frac{π}{3}

s u m m a r y :

x = k π - \frac{π}{6}

x = k π

x = 2 k π - \frac{π}{3}

x = 2 k π + \frac{2 π}{3}

Commented by EDWIN88 last updated on 29/Jan/21

n i c e

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