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Question Number 144936 by mathdanisur last updated on 30/Jun/21
Answered by mindispower last updated on 30/Jun/21
sec4(x)+csc4(x)⩾2cos2(x)sin2(x)=8(sin(2x))2⩾8sec6(y)+csc6(y)⩾2cos3(y)sin3(y)=16sin3(2y)⩾16sec8(z)+csc8(z)⩾2(cos(z)sin(z)4⩾32⇔⩾(8.16.32)13=16=ifonlyifsin(2x)=sin(2y)=sin(2z)=1⇒x=y=z=π4
Commented by mathdanisur last updated on 30/Jun/21
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