Question

sin^2 3x - cos^2 3x + 1/корень из 2 =0​

Answer 1

Ответ:

х = ±π/24 + πn/3 , n∈Z

Объяснение:

$\displaystyle \sin {}^{2} 3x - \cos {}^{2} 3x + \frac{1}{ \sqrt{2} } = 0 \\ - ( \cos {}^{2}3 x - \sin {}^{2} 3x) = - \frac{1}{ \sqrt{2} } \\ - \cos6x = - \frac{1}{ \sqrt{2} } \\ \cos6x = \frac{ \sqrt{2} }{2} \\ 6x = \pm \text{arccos} \frac{ \sqrt{2} }{2} + 2 \pi n \\ 6x = \pm \frac{ \pi}{4} + 2 \pi n|:6 \\ x = \pm \frac{ \pi}{24} + \frac{ \pi n}{3} ,n\in Z$