cos3x+1/2 akar 2=akar 2 0

cos 3x = akar dua - 1/2 akar 2
cos 3x = 1/2 akar 2
vos 3x = cos 30

* 3x = 30 + k.360
x = 10 + k.120

k=0 --- x = 10
k=1 --- x = 10 + 120 = 130
k=2 --- x = 10 + 240 = 250

* 3x = -30 + k360
x = -10 + k.120

k=0 --- x = -10 = 350
k= 1 --- x = -10 + 120 = 110
k=2 --- x = -10 + 240 = 230

[tex]cos3x + \frac{1}{2} \sqrt{2} = \sqrt{2} \\ cos3x = \sqrt{2} - \frac{1}{2} \sqrt{2} \\ cos3x = \frac{1}{2} \sqrt{2} \\ cos3x = cos 45°[/tex]

3x = 45° + 360°k
3x = 45° + 360°k
 x  = 15° + 120°k
⇒ k = 0, x = 15°
    k = 1, x = 135°
    k = 2, x = 255°

3x = -45° + 360°k
  x = -15° + 120°k
⇒ k = 1, x = 105°
    k = 2, x = 225°
    k = 3, x = 345°

Untuk syarat 0° < x < 360°, x = {15°, 105°, 135°, 225°, 255°, 345°}

Maap kalo salahh