Решите тригонометрические неравенство: 1) cos 2x < 0 2) sin x > 1/4

Предыдущий
вопрос Следующий
вопрос

1) cos 2x < 0

cos 2x <
cosine function is negative in the second and third quadrant
so, we have
π < 2x < 2π/3 or 4π/3 < 2x < 2π

Dividing by 2
π/2 < x < π/3 or 2π/3 < x < π

So the solution is
π/2 < x < π/3 or 2π/3 < x < π

2) sin x > 1/4

sin x > 1/
sine function is positive in the first and second quadrant
so, we have
0 < x < π/2 or π < x < 3π/2

So the solution is
0 < x < π/2 or π < x < 3π/2