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Solve the following equations:sinθ+3cosθ = 1

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प्रश्न

Solve the following equations:
`sin theta + sqrt(3) cos theta` = 1

योग
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उत्तर

Divide each term by 2

`1/2 sin theta + sqrt(3)/2 cos theta = 1/2`

`sin  pi/6 * sin theta + cos  pi/6 * cos theta = 1/2`

`cos theta * cos  pi/6 + sin theta * sin  pi/6 = 1/2`

`cos (theta - pi/6) = cos (pi/3)`

The general solution is

`theta - pi/6 = 2"n"pi +- pi/3`, n ∈ Z

θ = `2"n"pi +-  pi/3 + pi/6`, n ∈ Z

θ = `2"n"pi + pi/6 +- pi/3`, n ∈ Z

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 3: Trigonometry - Exercise 3.8 [पृष्ठ १३३]

APPEARS IN

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

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