Question

Find all value of x € ( 0 , π / 2 ) such that $\displaystyle{\frac{\sqrt{3}-1}{\sin x} +\frac{\sqrt{3}+1}{\cos x} =4\sqrt{2}}$

Answer 1

Answer:

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Answer 2

Question :--- Find all value of x € ( 0 , π / 2 ) such that (√3-1/sinx) + (√3+1/cosx) = 4√2 ..

Solution :----

→ (√3-1/sinx) + (√3+1/cosx) = 4√2

Taking LCM of LHS, we get,

→ [ (√3-1)*cosx + (√3+1)*sinx ] /sinx * cosx = 4√2

Taking Denominator RHS now, we get,

→ [ (√3-1)*cosx + (√3+1)*sinx ] = 4√2 * sinx * cosx

Dividing both sides by 2√2 now , we get,

→ [ (√3-1/2√2) * cosx + (√3+1/2√2)*sinx ] = 2 * sinx * cosx .

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Now, we know that,

[ → (√3-1/2√2) = sin15° = sin(π/12)

[ → (√3-1/2√2) = sin15° = sin(π/12)→ (√3+1/2√2) = cos15° = cos(π/12)

[ → (√3-1/2√2) = sin15° = sin(π/12)→ (√3+1/2√2) = cos15° = cos(π/12) → 2 * sinx * cosx = sin2x

[ → (√3-1/2√2) = sin15° = sin(π/12)→ (√3+1/2√2) = cos15° = cos(π/12) → 2 * sinx * cosx = sin2x → SinA*cosB + cosA*SinB = sin(A+B) ]

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using all these values in Question now, we get,

→ Sin(π/12) * cosx + cos(π/12)* sinx = sin2x

→ sin(x + π/12) = sin2x

→ 2x = nπ + (-1)^n ( x + π/12)

So, now putting n = 0, we get,

→ 2x = 0 + 1(x + π/12)

→ 2x = (x+π/12)

→ 2x - x = π/12

→ x = π/12 .

And, if n = 1 ,

→ 2x = π + (-1)(x + π/12)

→ 2x = π - x - π/12

→ 3x = 11π/12

→ x = 11π/36 .

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Hence, all value of x € ( 0 , π / 2 ) such that (√3-1/sinx) + (√3+1/cosx) = 4√2 are π/12 and 11π/36 ..