Me i've been on this problem for 20 mins

Hello from MrBillDoesMath!

  x < 3

Discussion:

Let's solve  (1/3)x + 2x  > 3x - 2. Note that "x" occurs in three terms. Let's combine them by bringing all x's to one side of the equation. Subtract (1/3)x from the both sides:

(1/3)x - (1/3)x + 2x > 3x - 2  - (1/3)x  or

2x > 3x - (1/3)x -  2.

Subtract 2x from both sides

2x - 2x > 3x - 2x - (1/3)x - 2 or

0 >  x - (1/3)x - 2.

Add 2 to both sides:

0 + 2 >  x - (1/3)x -2 + 2 or

2 >  x - (1/3)x or

2 > x(2/3).

Multiply both sides by 3/2 to get:

2 * (3/2) > x (2/3) * (3/2) or

3 >  x (1)

Thank you,

MrB

hope it you

step-by-step explanation:

sin(7pi/6 - pi/3)  

= sin(7pi/6 - 2pi/6)  

= sin(5pi/6)  

= sin(pi - pi/6)  

= sin(pi/6)  

= 1/2  

sin 7pi/6 - sin pi/3  

= sin (pi + pi/6) - sin pi/3  

= -sin pi/6 - sin pi/3  

= -1/2 - sqrt(3)/2  

= -(1+sqrt(3))2