First slide

Trigonometric equations

Question

If $\sin 3 α = 4 \sin α \sin (x + α) \sin (x - α)$ then

Easy

A

$x = n π \pm \frac{π}{3}, n \in z$
B

$x = n π \pm \frac{π}{6}, n \in z$
C

$x = n π \pm \frac{π}{2}, n \in z$
D

$x = 2 n π \pm \frac{π}{2}, n \in z$

Solution

$\sin 3 α = 4 \sin α \sin (x + α) \sin (x - α)$
$\Rightarrow 3 \sin α - 4 \sin^{3} α = 4 \sin α (\sin^{2} x - \sin^{2} α)$
$\Rightarrow 3 - 4 \sin^{2} α = 4 (\sin^{2} x - \sin^{2} α)$
$\Rightarrow 3 = 4 \sin^{2} x \Rightarrow \sin^{2} x = \frac{3}{4}$
$\Rightarrow \sin^{2} x = \sin^{2} \frac{π}{3} \Rightarrow x = n π \pm \frac{π}{3}, n \in z$

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