To solve the inequality 0 < tan(x) sec(x) < 1, we first need to understand the expressions involved. Recall that sec(x) = 1/cos(x) and tan(x) = sin(x)/cos(x). Thus, we can rewrite the product:
tan(x) sec(x) = tan(x) * (1/cos(x)) = (sin(x)/cos(x)) * (1/cos(x)) = sin(x) / cos^2(x).
Now, we want to find the solutions to the inequality:
- 1. First part: 0 < sin(x) / cos^2(x)
This means that sin(x) > 0 and cos^2(x) > 0. Since cos^2(x) > 0 for all x except where cos(x) = 0 (i.e., at odd multiples of π/2), we need to focus on where sin(x) > 0.
- 2. Second part: sin(x) / cos^2(x) < 1
Rearranging gives:
sin(x) < cos^2(x).
To find solutions for this inequality, we can analyze the graphs of sin(x) and cos^2(x) or solve it algebraically. However, since we are looking for solutions in the interval [0, 2], let’s check critical points:
- x = 0: sin(0) = 0, cos(0) = 1, thus 0 < 1.
- x = π/2: sin(π/2) = 1, cos(π/2) = 0, thus undefined.
- x = π: sin(π) = 0, thus 0 < 1.
- x = 3π/2: sin(3π/2) = -1, thus not in the solution set.
- x = 2: sin(2) > 0, cos(2) > 0.
Next, in intervals (0, π), (π, 2), check where sin(x) < cos^2(x) holds true:
From testing values, we find:
- On (0, π/2), both conditions are satisfied.
- On (π/2, π), while sin(x) is positive, the value exceeds cos^2(x).
- On (π, 2), the sine is ≤0, hence not applicable.
In conclusion, the solutions in the interval are:
- x ∈ (0, π/2)
Thus, the final result is:
- x ∈ (0, π/2)