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Solve the following equations:sin θ + cos θ = 2

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प्रश्न

Solve the following equations:
sin θ + cos θ = `sqrt(2)`

बेरीज
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उत्तर

Divide each term by `sqrt(2)`

`1/sqrt(2) sin theta + 1/sqrt(2) cos theta = sqrt(2)/sqrt(2)`

`sin pi/4 sin theta + cos pi/4 cos theta` = 1

`cos theta * cos  pi/4 + sin theta * sin  pi/4` = 1

`cos (theta - pi/4)` = 1

`cos (theta - pi/4)` = cos θ

The general solution is

`theta - pi/4` = 2nπ, n ∈ Z

θ = `2"n"pi + pi/4`, n ∈ Z

θ = `(8"n"pi + pi)/4`, n ∈ Z

θ = `(8"n" + 1)  pi/4`, n ∈ Z

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पाठ 3: Trigonometry - Exercise 3.8 [पृष्ठ १३३]

APPEARS IN

सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
पाठ 3 Trigonometry
Exercise 3.8 | Q 3. (vi) | पृष्ठ १३३

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