Question

Sin^4(x)+2sin(x-pi/3)sin(x+pi/3)+cos(x-pi/6)cos(x+pi/6)=0

Answer 1

Ответ на картинке внизу страницы

Answer 2

To solve the trigonometric equation $\sin^4(x) + 2\sin(x-\frac{\pi}{3})\sin(x+\frac{\pi}{3}) + \cos(x-\frac{\pi}{6})\cos(x+\frac{\pi}{6}) = 0$, we can simplify the trigonometric expressions and then use trigonometric identities to further simplify the equation.

First, let's simplify the trigonometric expressions:

1. $\sin(x-\frac{\pi}{3})\sin(x+\frac{\pi}{3})$: We can use the identity $\sin(A)\sin(B) = \frac{1}{2}[\cos(A-B) - \cos(A+B)]$: $\sin(x-\frac{\pi}{3})\sin(x+\frac{\pi}{3}) = \frac{1}{2}[\cos(x-\frac{\pi}{3} - (x+\frac{\pi}{3})) - \cos(x-\frac{\pi}{3} + (x+\frac{\pi}{3}))]$ $= \frac{1}{2}[\cos(x-\frac{\pi}{3} - x - \frac{\pi}{3}) - \cos(x+\frac{\pi}{3} - x + \frac{\pi}{3})]$ $= \frac{1}{2}[\cos(-\frac{2\pi}{3}) - \cos(\frac{2\pi}{3})]$ $= \frac{1}{2}[-\frac{1}{2} - \frac{1}{2}] = -\frac{1}{2}$

2. $\cos(x-\frac{\pi}{6})\cos(x+\frac{\pi}{6})$: We can use the identity $\cos(A)\cos(B) = \frac{1}{2}[\cos(A-B) + \cos(A+B)]$: $\cos(x-\frac{\pi}{6})\cos(x+\frac{\pi}{6}) = \frac{1}{2}[\cos(x-\frac{\pi}{6} - (x+\frac{\pi}{6})) + \cos(x-\frac{\pi}{6} + (x+\frac{\pi}{6}))]$ $= \frac{1}{2}[\cos(x-\frac{\pi}{6} - x - \frac{\pi}{6}) + \cos(x+\frac{\pi}{6} - x + \frac{\pi}{6})]$