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If x2+4+3sin(ax+b)−2x=0 has atleast one real solution, where a,b∈[0,2π], then the value of a+b can be |
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Answer» If x2+4+3sin(ax+b)−2x=0 has atleast one real solution, where a,b∈[0,2π], then the value of a+b can be |
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