Final answer:
The inequalities tanx > sinx, cosx > sinx, and −1 ≤ tanx ≤ 1 can be solved by considering the behaviors of the trigonometric functions in different quadrants within the interval [0, 2π]. Each inequality has specific intervals where it holds true.
Step-by-step explanation:
We're solving trigonometric inequalities within the interval [0, 2π].
a) tanx > sinx
The inequality tanx > sinx is satisfied when tanx is positive and greater than sinx, which occurs in the first and third quadrants. However, since sinx can never be greater than 1 and tanx can have values greater than 1, we also need to consider the intervals where sinx is less than 1 and tanx is positive, mainly the interval between 0 and π/2 and between π and 3π/2.
b) cosx > sinx
The inequality cosx > sinx is true when x is in the first quadrant (π/4 to be precise, where cosx equals sinx) and in the fourth quadrant, before sinx becomes positive and larger than cosx.
c) −1 ≤ tanx ≤ 1
The inequality −1 ≤ tanx ≤ 1 holds true for values of x where tan is within the range of -1 to 1. This can happen around the angle 0, π, and 2π, where tan approaches 0, and in intervals around π/4 and 3π/4 where the tan function has values of 1 and -1, respectively.
In summary, no single option is universally correct without further refinement of where in the specified range they hold, as these functions have different behavior in different quadrants.