Question

1) SIN⁴2x+COS⁴2x=SIN2x·COS2x
2)6·SIN²x+SINx·COSx-COS²x=2
Решить, используя тригонометрические формулы

Answer 1

Sin⁴2x + Cos⁴2x = (sin²2х+соs²2x)² - 2sin²2xcos²2x = 1 - 2sin²2xcos²2x = 1 - sin²4x 1 - sin²4x = 1/2 sin4x 2sin²4x + sin4x - 2 = 0 2t² + t - 2 = 0 t1,2 = 0,5; -1 sin4x = -1 x = 3П/8 + Пк/2 sin4x = 0,5 x = (-1)ⁿπ/12 + πк/4