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Find the general solutions of the following equation: sin θ - cos θ = 1 - Mathematics and Statistics

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प्रश्न

Find the general solutions of the following equation:

sin θ - cos θ = 1

योग
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उत्तर

sin θ − cos θ = 1

∴ cos θ − sin θ = −1

∴ (1) cos θ − (1) sin θ = −1

`sqrt((1)^2 + (1)^2) = sqrt(1 + 1) = sqrt2`

dividing b. s. by `sqrt2`

∴ `1/sqrt2 costheta - 1/sqrt2 sintheta = -1/sqrt2`

∴ `cos  pi/4 costheta - sin  pi/4 sintheta = - cos  pi/4`

∴ `cos"A" cos"B" - sin"A" sin"B" = cos"(A + B)"`

∴ `cos(theta + pi/4) = cos (pi - pi/4)  ...(∵ - costheta = cos(pi - theta))`

∴ `cos (theta + pi/4) = cos  (3pi)/4`

cos θ = cos α ⇒ θ = 2n π ± α, n ∈ 2

∴ `theta + pi/4 = 2npi +- (3pi)/4, n ∈ 2`

∴ `theta = 2npi +- (3pi)/4 - pi/4, n ∈ 2`

∴ `theta = 2n pi + (3pi)/4 - pi/4 or theta = 2npi - (3pi)/4 - pi/4, n ∈ 2`

∴ `theta = 2npi + (2pi)/4 or theta = 2npi - (4pi)/4 - pi/4,  n ∈ 2`

∴ `theta = 2npi + pi/2 or theta = 2npi - pi n ∈ 2`

∴ These are required general solutions.

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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 4.3 | पृष्ठ १०९

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