Помогите, пожалуйста 1) 2cos^2 a = 1 + -/2 – cos 2x 2) cos^2 6x + sin^2 3x – 1 = 0

Question

Помогите, пожалуйста
1) 2cos^2 a = 1 + -/2 – cos 2x
2) cos^2 6x + sin^2 3x – 1 = 0

Comments

в 1) после 1 + ....?

Answer 1

2) sin^2 3x = 1 – cos^2 6x

sin^2 6x – sin^2 3x = 0

4sin^2 3x * cos^2 3x – sin^2 3x = 0

sin^2 3x*(4cos^2 3x - 1) = 0

sin^2 3x = 0    cos^2 3x =1/4

По формулам понижения степени:

(1 - cos 6x)/2 = 0              (1 + cos 6x)/2 = ¼

co6x = 1                             cos6x = - ½

x = pi/3 * n                        6x = ±(2pi/3) +2pi n

                                              x = ±pi/4 + pi/3 *n