數學挑戰題 (2) --- 三角函數不等式 (20分)

zerothpoint's answer to 1) must be wrong since (-π < x < π) given by question, therefore, 3π/2 or 2π have exceeded the limit.
(1)
1 + sin x ≧ 2 cos² x (-π < x < π)
=> 1 + sin x ≧ 2 (1 - sin² x)
=> 1 + sin x ≧ 2 - 2 sin² x
=> 2 sin² x - 2 +1 + sin x ≧ 0
=> 2 sin² x + sin x -1 ≧ 0
=> (2 sin x -1) (sin x +1) ≧ 0
Since (2 sin x -1) (sin x +1) ≧ 0, therefore, either both (2 sin x -1) and (sin x +1) ≧ 0 or both (2 sin x -1) and (sin x +1) ≦ 0

Scenario 1: Both (2 sin x -1) and (sin x +1) ≧ 0
(2 sin x -1) ≧ 0
=> sin x ≧ 1/2
=> π/6 ≦ x ≦ 5π/6 .....(a)

(sin x +1) ≧ 0
=> sin x ≧ -1
Since -π < x < π, i.e., it lies on all quadrants,and any points are valid between -π < x < π .....(b)

Combining (a) and (b), π/6 ≦ x ≦ 5π/6

Scenario 2: Both (2 sin x -1) and (sin x +1) ≦ 0
(2 sin x -1) ≦ 0
=> sin x ≦ 1/2
=> π > x ≧ 5π/6 or -π < x ≦ π/6 .....(c)

(sin x +1) ≦ 0
=> sin x ≦ -1
Since -π < x < π, ie, only one point is valid, ie, sin x = -1 means x = -π/2
....(d)
Thus the only solution to Scenario is x = -π/2 after comparing (c) and (d).

Hence, combining Scenarios 1 and 2:
π/6 ≦ x ≦ 5π/6 or x = -π/2 (Answer.)

(2) 2 cos² x > cos x and sin x > 0 given 0 ≦x< 2π
Since sin x > 0, x is confined in quadrants I and II, ie, 0 < x < π

2 cos² x > cos x
=> 2 cos² x - cos x > 0
=> (2 cos x - 1) cos x > 0


Since (2 cos x - 1) cos x > 0, therefore, either both (2 cos x - 1) and cos x > 0 or both (2 cos x - 1) < 0 and cos x < 0

Scenario 3: Both (2 cos x - 1) and cos x > 0
(2 cos x - 1) > 0
=> cos x > 1/2
=> 0 < x < π/3 .....(e)

cos x > 0
=> 0 < x < π/2 .....(f)
combining (e) and (f), 0 < x < π/3

Scenario 4: Both (2 cos x - 1) and cos x < 0
(2 cos x - 1) < 0
=> cos x < 1/2
=> π/3 < x < π ..... (g)

cos x < 0
=> π/2 < x < π ......(h)
combining (g) and (h), π/2 < x < π

By combining Scenarios 3 and 4, 0 < x < π/3 or π/2 < x < π (Answer)

Q1
1 + sin x ≧2 ( cos x )^2
1 + sin x ≧2 (1 - (sin x)^2)
2(sin x)^2 + sin x - 1 ≧ 0
(2sin x - 1)(sin x + 1) ≧ 0
-1 <= sin x <= 1/2
(0 <= x <= π/3) or (2π/3 <=x < 3π/2) or ( 3π/2 < x <= 2π)

Q2
The inequality sin x > 0 implies 0 < x < π
So, we have
2 (cos x)^2 > cos x
cos x (2cos x - 1) > 0
cos x < 0 or cos x > 1/2
(0 <= x <= π/3) or (π/2 < x < π)