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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`]
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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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