8 sin^2x + cos 2x + 1 = 0

Предыдущий
вопрос Следующий
вопрос

To solve this equation, we need to use trigonometric identities.
We know that cos 2x = 2cos^2x - 1, so we can substitute it into the given equation:

8sin^2x + 2cos^2x - 1 + 1 = 0
8sin^2x + 2cos^2x = 0
8sin^2x + 2(1 - sin^2x) = 0
8sin^2x + 2 - 2sin^2x = 0
6sin^2x + 2 = 0
6sin^2x = -2
sin^2x = -2/6
sin^2x = -1/3

Since sin^2x cannot be negative, there are no solutions to this equation.