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Find the general solution of the equation (3-1)cosθ+(3+1)sinθ = 2 [Hint: Put 3-1 = r sinα, 3+1 = r cosα which gives tanα = tan(π4-π6) α = π12] - Mathematics

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प्रश्न

Find the general solution of the equation `(sqrt(3) - 1) costheta + (sqrt(3) + 1) sin theta` = 2

[Hint: Put `sqrt(3) - 1` = r sinα, `sqrt(3) + 1` = r cosα which gives tanα = `tan(pi/4 - pi/6)` α = `pi/12`]

योग
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उत्तर

Given that: `(sqrt(3) - 1) costheta + (sqrt(3) + 1) sin theta` = 2

Put `sqrt(3) - 1` = r sinα, `sqrt(3) + 1` = r cosα

By squaring and adding, we get

r2 = `3 + 1 - 2sqrt(3) + 3 + 1 + 2sqrt(3)`

⇒ r2 = 8

⇒ r = `+-  2sqrt(2)`

Now the given equation can be written as

rsinα cosθ + rcosα sinθ = 2

⇒ r(sinα cosθ + cosα sinθ) = 2

⇒ `2sqrt(2) sin(alpha + theta)` = 2

⇒ `sin(alpha + theta) = 2/(2sqrt(2)) = 1/sqrt(2)`

⇒ `sin(alpha + theta) = sin  pi/4`

∴ α + θ = `npi + (-1)^n * pi/4`  .....(i)

Now  `(r sin alpha)/(r cos alpha) = (sqrt(3) - 1)/(sqrt(3) + 1)`

⇒ tanα = `(tan  pi/3 - tan  pi/4)/(1 + tan  pi/4 * tan  pi/3)`

⇒ tanα = `tan(pi/3 - pi/4)`

⇒ tanα = `tan  pi/12`

∴ α = `pi/12`

Putting the value of α in equation (i) we get

`pi/12 + theta = npi + (-1)^n * pi/4`

∴ θ = `npi + (-1)^n * pi/4 - pi/12`

Hence, the general solution of the given equation is θ = `npi + (-1)^n * pi/4 - pi/12`, n ∈ Z. 

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अध्याय 3: Trigonometric Functions - Exercise [पृष्ठ ५५]

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एनसीईआरटी एक्झांप्लर Mathematics [English] Class 11
अध्याय 3 Trigonometric Functions
Exercise | Q 29 | पृष्ठ ५५

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