x=±0,5arccos(7/9)+kπ, k={1; 2}
Объяснение:
24 tg²x - 9 sin²x = 2
24 sin²x/cos²x - 9 sin²x = 2
0≠cos²x=t⇒0<t≤1
sin²x=1-cos²x=1-t
24(1-t)/t-9(1-t)=2
t[24(1-t)/t-9(1-t)]=2t
24(1-t)-9t(1-t)=2t
24-24t-9t+9t²-2t=0
9t²-35t+24=0
D=1225-864=361=19²
t₁=(35-19)/18=16/18=8/9
t₂=(35+19)/18=3>1
(1+cos2x)/2=cos²x=t=8/9
1+cos2x=16/9
cos2x=7/9
2x=±arccos(7/9)+2kπ
x=±0,5arccos(7/9)+kπ, k∈Z
0<arccos(7/9)<π/4
0<0,5arccos(7/9)<π/8, x∈[(3π)/4; (9π)/4]⇒x=±0,5arccos(7/9)+kπ, k={1; 2}
cos5x + cosx + 2cos2x = 0
2cos(5x+x/2)cos(5x-x/2) + 2cos2x = 0
2cos(6x/2)cos(4x/2) + 2cos2x = 0
2cos3x × cos2x + 2cos2x = 0
2cos2x × (cos3x + 1) = 0 | : 2
cos2x × (cos3x + 1) = 0
cos2x = 0 или cos3x + 1 = 0
2x = π/2 + πn cos3x = -1
x₁ = π/2 × 1/2 + πn × 1/2 3x = π + 2πn
x₁ = π/4 + πn/2, n∈Z x₂ = π × 1/3 + 2πn × 1/3
x₂ = π/3 + 2πn/3, n∈Z
ответ: x₁ = π/4 + πn/2, n∈Z
x₂ = π/3 + 2πn/3, n∈Z
