If 2 s e c 2 x tan 2 y − tan y + 2 1 ≤ 1 , how many values of x ∈ [ 0 , 2 π ] can occur?

The square root is at least 1/2 and sec 2 x ≥ 1 , so equality is required. Thus sec 2 x = 1 , giving x = 0 , π , 2 π .

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MathematicsTrigonometric Ratios & IdentitiesMaximum & Minimum ValuesMedium3 minQB

MathematicsMediumsingle choice

If $2^{s e c^{2} x} tan^{2} y - tan y + \frac{1}{2} \leq 1$ , how many values of $x \in [0, 2 π]$ can occur?

Answer:

Solution:

The square root is at least $1/2$ and $sec^{2} x \geq 1$ , so equality is required. Thus $sec^{2} x = 1$ , giving $x = 0, π, 2 π$ .

Stream:JEESubject:MathematicsTopic:Trigonometric Ratios & IdentitiesSubtopic:Maximum & Minimum Values

⏱ 3mℹ️ Source: QB

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