solve 2sinx+√3<0 for 0<x<2π
sinx < -√3/2
consider sinx = -√3/2
by the CAST rule and knowing that sin 30° = +√3/2
x = 210° or x = 330°
also, you should know the general shape of the sine curve.
from the above the curve would be below -√3/2 between 210° and 330°
or in radians:
7π/6 < x < 11π/6
verification:
http://www.wolframalpha.com/input/?i=sinx++%3C+-%E2%88%9A3%2F2+from+0+to+2%CF%80
To solve the inequality 2sin(x) + √3 < 0 for 0 < x < 2π, we can follow these steps:
Step 1: Subtract √3 from both sides of the inequality:
2sin(x) < -√3
Step 2: Divide both sides of the inequality by 2:
sin(x) < -√3/2
Step 3: Determine the reference angle for -√3/2:
The reference angle for sin(x) = -√3/2 is π/3.
Step 4: Determine the solutions in the given interval:
Since sin(x) < 0 in the given interval 0 < x < 2π, this means that the solutions lie in the third and fourth quadrants.
In the third quadrant, the angle is between π and 3π/2. Therefore, one solution is x = π + π/3 = 4π/3.
In the fourth quadrant, the angle is between 3π/2 and 2π. Therefore, another solution is x = 2π - π/3 = 5π/3.
Step 5: Write the final solution:
The solution to the inequality 2sin(x) + √3 < 0, for 0 < x < 2π, is x = 4π/3 and x = 5π/3.
Therefore, for x values between 0 and 2π, the solution to the inequality is x = 4π/3 and x = 5π/3.