First slide

Trigonometric equations

Question

$Solution of the equation 3 \sqrt{3} \sin^{3} x + \cos^{3} x + 3 \sqrt{3} \sin x \cos x = 1$

Difficult

A

$x = (n π + {(- 1)}^{n} \frac{π}{6}) - \frac{π}{6}, n \in I$
B

$x = (2 n + 1) \frac{π}{6}, n \in I$
C

$x = 2 n π + \frac{π}{3}, n \in I$
D

$x = n π + {(- 1)}^{n} \frac{π}{3}, n \in I$

Solution

$\begin{array}{l} It is in the form of a^{3} + b^{3} + c^{3} - 3 a b c = 0 \\ (a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - c a) = 0, (\sqrt{3} \sin x)^{3} + (\cos x)^{3} + (- 1)^{3} - 3 (\sqrt{3} \sin x) (\cos x) (- 1) = 0 \end{array}$
$\begin{array}{l} \sqrt{3} \sin x + \cos x - 1 = 0 \\ \sqrt{3} \sin x + \cos x = 1, divided by 2 \\ \sin (x + \frac{π}{6}) = \sin \frac{π}{6} \\ x + \frac{π}{6} = n π + (- 1)^{n} \frac{π}{6}, n \in z \end{array}$

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