 # Find the value of a for which 4cosec2  ( π (a + x)) +a2 - 4a = 0 has a real solution. Priyansh Bajaj AskiitiansExpert-IITD
30 Points
11 years ago

Dear Nirabhra,

Solution:- 4cosec2(n(a+x)) + a2 - 4a =0

[4cosec2(n(a+x)) -4] + (a2 - 4a + 4) =0

4[cosec2(n(a+x)) -1] + (a-2)2 =0

[2cot(n(a+x))]2 + (a-2)2 =0 which means a =2 and cot(n(a+x)) =0

Ans: a= 2

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Priyansh Bajaj

11 years ago

Dear Nirabhra,

Solution:- 4cosec2(n(a+x)) + a2 - 4a =0

[4cosec2(n(a+x)) -4] + (a2 - 4a + 4) =0

4[cosec2(n(a+x)) -1] + (a-2)2 =0

[2cot(n(a+x))]2 + (a-2)2 =0 which means a =2 and cot(n(a+x)) =0

4 years ago
- 4a = 02 (π (a+x) ) + a24 cosec

4 cosec 2 (π (a+x) ) =- a2 + 4a

L .H .S is always greater than 4 so for real solution

-a2 + 4a ≥4

or (a-2)2≤0

so only a=2 is possible