Ответ(ы) на вопрос:
Гость 2sin^2 x = sin x + 1;
2 sin^2 x - sin x - 1 = 0;
D = 1 + 8 = 9 = 3^2;
sin x = 1; x = π/ 2 + 2 πk; k∈Z;
sin x = - 1/2; x = (π/6)^(k+1) + πk; k∈ Z.
2 sin 2x - 3 sin^2 x = 1;
4 sinx cos x - 3 sin^2 x = sin^2 x + cos^2 x;
- 4 sin^2 x + 4 sinx cos x - cos^2 x = 0;
4 sin^2 x - 4 sin x cos x + cos^2 x= 0; /: cos^2 x ≠ 0;
4 tg^2 x - 4 tg x + 1 = 0;
D= 16 - 4*4 = 0;
tg x = 2;
x = arctg2 + πk; k∈Z.
sin(5π/6 + x/2) = - √3/2;
1) 5π /6 + x/2 = -π /3 + 2πk;
x/2 = - π /3 - 5π /6 + 2 π k;
x/2= - 7π /6 + 2π k; /*2
x = - 7π /3 + 4 π k;
x = - π /3 + 4π k; k∈Z
2) 5π /6 + x/2 = - 2π /3 + 2π k;
x / 2 = - 2 π /3 - 5π /6 + 2 π k;
x/2 = - 9 π /6 + 2π k;
x/2 = - 3π /2 + 2 π k;
x/2 = π /2 + 2π k /*2
x = π + 4 π k; k∈Z
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