cotx = - rad3 / 3
does x= 2pi/3 and 5pi/3?
since cotx < 0 in QII and QIV, you are correct.
To determine if x = 2pi/3 and 5pi/3 are solutions to cotx = -sqrt(3)/3, we can substitute these values into the equation and check if they satisfy the equation.
First, let's substitute x = 2pi/3 into cotx = -sqrt(3)/3:
cot(2pi/3) = -sqrt(3)/3
To find the cotangent of an angle, we can use the reciprocal of the tangent function. So, cot(2pi/3) is equal to 1/tan(2pi/3).
tan(2pi/3) is equal to sqrt(3) (since tan(2pi/3) = sin(2pi/3) / cos(2pi/3), and sin(2pi/3) = sqrt(3)/2 and cos(2pi/3) = -1/2).
Therefore, cot(2pi/3) = 1/(sqrt(3)) = sqrt(3)/3
Since sqrt(3)/3 is equal to -sqrt(3)/3, x = 2pi/3 satisfies the equation cotx = -sqrt(3)/3.
Now, let's substitute x = 5pi/3 into cotx = -sqrt(3)/3:
cot(5pi/3) = -sqrt(3)/3
Again, using the reciprocal of the tangent function, we have cot(5pi/3) = 1/tan(5pi/3).
tan(5pi/3) is equal to -sqrt(3) (since tan(5pi/3) = sin(5pi/3) / cos(5pi/3), and sin(5pi/3) = -sqrt(3)/2 and cos(5pi/3) = -1/2).
Therefore, cot(5pi/3) = 1/(-sqrt(3)) = -sqrt(3)/3
Since -sqrt(3)/3 is equal to -sqrt(3)/3, x = 5pi/3 also satisfies the equation cotx = -sqrt(3)/3.
In conclusion, x = 2pi/3 and x = 5pi/3 are both solutions to the equation cotx = -sqrt(3)/3.