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Find the general solutions of the following equation: tan2θ=3

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प्रश्न

Find the general solutions of the following equation:

`tan^2 theta = 3`

योग
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उत्तर

The general solution of tan2θ = tan2α is θ = nπ ± α, n ∈ Z.

Now, tan2 θ = 3 = `(sqrt3)^2`

∴ tan2 θ = `(tan  pi/3)^2    ....[because "tan" pi/3 = sqrt3]`

∴ tan2 θ = `tan^2  (pi)/3`

∴ the required general solution is

∴ θ = `"n"pi +- (pi)/3,` n ∈ Z.

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 3: Trigonometric Functions - Miscellaneous exercise 3 [पृष्ठ १०९]

APPEARS IN

बालभारती Mathematics and Statistics 1 (Arts and Science) [English] Standard 12 Maharashtra State Board
अध्याय 3 Trigonometric Functions
Miscellaneous exercise 3 | Q 4.2 | पृष्ठ १०९

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