# kow to solve the inequality trigonometry for 2 sin x cos x - sin x + 2 cos x -1 <0 and  -180 <= x <= 180

giorgiana1976 | Student

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

and

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

2cos x>1

cos x>1/2

x > pi/3 or x> 2pi/3

We'll discuss the 2nd case:

2)

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.

and

2cos x - 1<0 => x < pi/3

The intervals of admissible values for x are: (0 , pi/3)U(pi/3 , pi).

nang1350 | Student

NOTE 2.

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).

nang1350 | Student

NOTE

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.

nang1350 | Student

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)

Nang1350

nang1350 | Student

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.