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First, we'll factorize by 2cos x the 1st and the 3rd terms:
2 cos x(sin x + 1) - (sin x + 1)<0
We'll factorize by (sin x + 1):
(sin x + 1)*(2cos x - 1) < 0
A product is negative if the factors have opposite signs.
We'll discuss 2 cases:
1) sin x + 1 < 0
2cos x - 1>0
sin x < -1
This is impossible since the values of sines function cannot be smaller than -1.
2cos x - 1>0
x > pi/3 or x> 2pi/3
We'll discuss the 2nd case:
sin x + 1 > 0<=> sin x>-1 => x> pi
This is impossible since x is in the range [0,180] and cannot be larger than 180.
2cos x - 1<0 => x < pi/3
The intervals of admissible values for x are: (0 , pi/3)U(pi/3 , pi).
Now, let's see what is the solution set within the period (-pi, pi).
a. The function u = sin x + 1 is always positive.
b. The function v = 2cos x -1 is negative within the interval (-pi, -pi/3)
Check with x = -pi/2 ; cos x = -1 , v = -2 = 1 = -1, negative
c. The function v = 2cos x -1 is positive within the interval (-pi/3, pi/3)
Check with x = 0; cos x = 1, v = 2 -1 = 1, positive
d. The function v = 2cos x is also negative within interval (pi/3, pi)
Check with x = pi/2 ; cos x = 0, v = - 1, negative
Conclusion. The solution set of the trig inequality within the period (-pi, pi) is the intervals (-pi, -pi/3) and (pi/3, pi).
1. The trig function u = sin x + 1 is always positive within the period (0, 2Pi).
Check: When x = 0; sin x = 0; u = 1 > 0
When x = Pi/2; sinx = 1; u = 2 > 0
When x = Pi; sin x = 0; u = 1 > 0
2. The sign status of the product u*v is the sign status p of the trig function v = 2cos x -1.
Within interval (0, Pi/3), v is positive
Within the interval (Pi/3, 5Pi/3) v is negative.
Within the interval (5pi/3, 2Pi), v is positive.
Conclusion. The solution set of the inequality is the interval (pi/3, 5pi/3) where the product u*v is negative.
Sorry, Please correct one mistake in the answer above.
Solve v = (2cos x - 1) = 0 --> cos x = 1/2 --> x = Pi/3 and x = 5Pi/3
The solution set of the inequality is the interval (Pi/3 , 5Pi/3)
a. First, transform the trig inequality into a product:
F = (sin x + 1)(2cos x -1) < 0
b. Solve u = (sin x +1) = 0 and v = (2 cos x - 1) = 0
sin x = 1 --> x = 3Pi/2; cos x = 1/2 --> x = Pi/3 and x = 4Pi/3
c. Set up a Sign Table: Variation of u and v within period 2Pi
x | 0 Pi/3 Pi 3Pi/2 4Pi/3 2Pi
u | + + + 0 + +
v | + 0 - - - 0 +
F | No yes yes yes No
The sign of F is the sign of the product u*v
The solution set of the inequality is the interval ( Pi/3, 4Pi/3)
Note 1: To solve a trig inequality, transform it into a product of basic trig inequalities. Then, set up a Sign Table that shows all the solutions sets of these basic inequalities. The combined solution set can be easily seen as the sign of a product
Note 2. Don't try to discuss the signs of the 2 binomials since it will lead to confusion and error.
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