e|sinx|+e−|sinx|+4a=0 will have exactly four different solutions in [0,2π] if
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Let e|sinx|=tt∈[1,e]∴t+1t+4a=0⇒t2+4at+1=0.∴ This quadratic equation has 2 distinct roots in [1,e] as x∈ [0,2π]⇒16a2−4>0, 1+4a+1≥0, e2+4at+1≥0, 1<−2a<e|a|>12, a≥−12, a≥−1−e24e, −e2<a<−12∴ No value satisfy the above inequalities simultaneously.