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If sin-1(1 - x) - 2 sin-1x = π2, then find the value of x. - Mathematics and Statistics

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प्रश्न

If sin-1(1 - x) - 2 sin-1x =  `pi/2`, then find the value of x.

योग
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उत्तर

sin-1(1 - x) - 2 sin-1x =  `pi/2`

∴ sin-1(1 - x) = `pi/2 + 2 sin^-1 "x"`

∴ 1 - x = sin `(pi/2 + 2 sin^-1"x")`

∴ 1 - x = cos (2 sin-1x) ....`[ ∵ sin (pi/2 + theta) = cos theta]`

∴ 1 - x = 1 - 2[sin(sin-1 x)]2   ....[∵ cos 2θ = 1 - 2 sin2θ]

∴ 1 - x = 1 - 2x2

∴ 2x2 - x = 0

∴ x(2x - 1) = 0

∴ x = 0  or x = `1/2`

When x = `1/2`

LHS = `sin^-1 (1 - 1/2) - 2 sin^-1 (1/2)`

`= sin^-1 (1/2) - 2 sin^-1 (1/2)`

`= - sin^-1 (1/2)`

`= - sin^-1 (sin  pi/6)`

`= - pi/6 ne pi/2`

∴ `"x" ne 1/2`

Hence, x = 0.

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अध्याय 3: Trigonometric Functions - Miscellaneous exercise 3 [पृष्ठ ११०]

APPEARS IN

बालभारती Mathematics and Statistics 1 (Arts and Science) [English] Standard 12 Maharashtra State Board
अध्याय 3 Trigonometric Functions
Miscellaneous exercise 3 | Q 28 | पृष्ठ ११०

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