Question

Решить уравнение:
sin^2 2x-cos 2x-1=0

Answer 1

Ответ:

$x_{1} = \frac{\pi}{4} + \frac{\pi \: n}{2} \\ x_{2} = \pi \: n$

n€Z

Объяснение:

${sin}^{2} 2x - cos2x - 1 = 0$

основное тригонометрическое тождество:

${sin}^{2} \alpha + {cos}^{2} \alpha = 1 \\ {sin}^{2} \alpha = 1 - {cos}^{2} \alpha$

$(1 - {cos}^{2}2x) - cos2x - 1 = 0 \\ - {cos}^{2} 2x - cos2x = 0 \\ - cos2x \times (cos2x + 1) = 0 \\ - cos2x = 0 \: \: ili \: \: cos2x + 1 = 0$

простейшие тригонометрические уравнения

$cos2x = 0 \\ 2x = \frac{\pi}{2} + \pi \: n$

n €Z

$x = ( \frac{\pi}{2} + \pi \: n) \div 2 \\ x = \frac{\pi}{4} + \frac{\pi \: n}{2}$

$cos2x + 1 = 0 \\ cos2x = - 1 \\ 2x =2\pi \: n \\ x = \pi \: n$