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प्रश्न
The general solution of sec θ = `sqrt2` is
विकल्प
2nπ ± `pi/3`, n ∈ Z
2nπ ± `pi/6`, n ∈ Z
nπ ± `pi/2`, n ∈ Z
2nπ ± `pi/4`, n ∈ Z
MCQ
उत्तर
2nπ ± `pi/4`, n ∈ Z
Explanation:
sec θ = `sqrt2`
`= cos theta = 1/sqrt2`
= cos θ = cos `pi/4`
= θ = 2nπ ± `pi/4`, n ∈ Z ....[∵ cos θ = cos α ⇒ θ = 2nπ ± α]
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Trigonometric Equations and Their Solutions
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