Let A= {x∈(0,π)-{π/2} ∶ 〖log〗_(2/π) |sinx|+〖log〗_(2/π) |cosx|=2} and B = {x≥0∶ √x (√x- 4)- 3|√x-2|+

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Let A = {x∈(0,π)-{π/2} : log₂/π | sinx| + log₂/π |cosx|=2} and B = {x≥0:√x(√x-4)-3|√x-2|+6=0}. Then Let A = {x∈(0,π)-{π/2} : log₂/π | sinx| + log₂/π |cosx|=2} and B = {x≥0:√x(√x-4)-3|√x-2|+6=0}. Then
Let A= {x∈(0,π)-{π/2} ∶ 〖log〗_(2/π) |sinx|+ 〖log〗_(2/π) |cosx|=2} and B = {x≥0∶ √x (√x- 4)- 3|√ Let A= {x∈(0,π)-{π/2} ∶ 〖log〗_(2/π) |sinx|+ 〖log〗_(2/π) |cosx|=2} and B = {x≥0∶ √x (√x- 4)- 3|√
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