NC Ремить уравнения 9) 48R²x + 11 cosx-1=0 6) STRIX + STUX-COSX-QCOS²x=0 6) 2005+4x+COS6x-1=0

a) 4sin²x + 11cosx - 1 = 0 4(1-cos²x) + 11cosx - 1 = 0 4 - 4cos²x + 11cosx - 1 = 0 -4cos²x + 11cosx + 3 = 0 4cos²x - 11cosx - 3 = 0 D = 121 - 4*4*(-3) = 121 + 48 = 169 cosx = (11 ± 13)/8

• cosx = (11 + 13)/8 = 24/8 = 3 (не подходит, так как cosx ∈ [-1, 1])
• cosx = (11 - 13)/8 = -2/8 = -1/4 x = ±arccos(-1/4) + 2πk, k ∈ Z

б) sin²x + sinx \(\cdot\) cosx - 2cos²x = 0 tg²x + tgx - 2 = 0 D = 1 - 4*1*(-2) = 1 + 8 = 9

• tgx = (-1 + 3)/2 = 1 x = \(\frac{\pi}{4}\) + \(\pi\)k, k ∈ Z
• tgx = (-1 - 3)/2 = -2 x = arctg(-2) + \(\pi\)k, k ∈ Z

в) 2cos²4x + cos6x - 1 = 0 cos8x + cos6x = 0 2cos7x \(\cdot\) cosx = 0

• cos7x = 0 7x = \(\frac{\pi}{2}\) + \(\pi\)k, k ∈ Z x = \(\frac{\pi}{14}\) + \(\frac{\pi}{7}\)k, k ∈ Z
• cosx = 0 x = \(\frac{\pi}{2}\) + \(\pi\)k, k ∈ Z

Ответ: a) x = ±arccos(-1/4) + 2πk, k ∈ Z; б) x = \(\frac{\pi}{4}\) + \(\pi\)k, k ∈ Z, x = arctg(-2) + \(\pi\)k, k ∈ Z; в) x = \(\frac{\pi}{14}\) + \(\frac{\pi}{7}\)k, k ∈ Z, x = \(\frac{\pi}{2}\) + \(\pi\)k, k ∈ Z