Find the value of a for which the eqn

4 cosec ² (π (a+x) ) + a² - 4a = 0 has a real solution.

Dear Sanchit

4 cosec ² (π (a+x) ) + a² - 4a = 0

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

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