For what intervals of x-x-0- - x- 0- 4-

For finding the region where x gt;0 ∀ x∈ (0, 4π), for that let's draw the graph of x. Thus, from observing the graph, we can say x gt;0 for x∈ (0, π/2)∪(3 ...

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For what inte...

Question

For what intervals of $x;$
$sec x > 0, \forall x \in (0, 4 π) ?$

x \in (7 π / 2, 4 π)

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x \in (π / 2, 3 π / 2)

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x \in (3 π / 2, 5 π / 2)

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x \in (0, π / 2)

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Solution

The correct option is D $x \in (0, π / 2)$
For finding the region where $sec x > 0 \forall x \in (0, 4 π),$ for that let's draw the graph of $sec x .$

Thus, from observing the graph, we can say $sec x > 0$ for
$x \in (0, π / 2) \cup (3 π / 2, 5 π / 2) \cup (7 π / 2, 4 π)$

Suggest Corrections

For what intervals of x; secx>0, ∀ x∈(0,4π)?

The correct option is D x∈(0,π/2) For finding the region where secx>0 ∀ x∈(0,4π), for that let's draw the graph of secx. Thus, from observing the graph, we can say secx>0 for x∈(0,π/2)∪(3π/2,5π/2)∪(7π/2,4π)

For what intervals of $x;$
$sec x > 0, \forall x \in (0, 4 π) ?$

The correct option is D $x \in (0, π / 2)$
For finding the region where $sec x > 0 \forall x \in (0, 4 π),$ for that let's draw the graph of $sec x .$

Thus, from observing the graph, we can say $sec x > 0$ for
$x \in (0, π / 2) \cup (3 π / 2, 5 π / 2) \cup (7 π / 2, 4 π)$