Solve 1/2cot(x)≥1/2 over 0≤x≤2pi Solving a Trigonometric Inequality











The answer is B.
The second one.
0<x≤pi/4 and pi≤x≤5pi/4

answer: about 22.67

step-by-step explanation:

no oblique asymptote

step-by-step explanation:

to find the oblique asymptote of f(x)

here numerator has degree 1 and denomintor degree 2

degree of numerator < degree of denominator

hence there is no oblique asymptote

when the degree of the denominator > that of the numerator, we saw that the value in the denominator got so much bigger, so quickly,   giving us a horizontal asymptote of the x-axis. or we can write

f(x) = a polynomial of x + a proper fraction.

in this case only we have oblique asymptotes but not when numerator has less degree.

step-by-step explanation:

this can be solved first with pythagorean theorem, which tells us to square both legs of the triangle and sum it. this will be equal to the square of the hypotenuse.

the hypotenuse is the side "opposite" the 90 degree angle, which is x.

so we can write using pythagorean theorem,

simplifying:

to further simplify this radical, we need to use the property  

thus