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SOLUTION
Write out the given information
[tex]\begin{gathered} \sin x+\cos x=\frac{5}{4} \\ \\ \sin 2x=2\sin x\cos x \end{gathered}[/tex]From the expression given
[tex]\begin{gathered} \sin x+\cos x=\frac{5}{4} \\ \text{Square both sides } \\ (\sin x+\cos x)^2=(\frac{5}{4})^2 \end{gathered}[/tex]Then, we have
[tex]\begin{gathered} \sin ^2x+2\sin x\cos x+\cos ^2x=\frac{25}{16} \\ \text{rearrange } \\ \sin ^2x+\cos ^2x+2\sin x\cos x=\frac{25}{16} \end{gathered}[/tex]Recall that
[tex]\begin{gathered} \sin ^2x+\cos ^2x=1 \\ \text{and} \\ \sin 2x=2\sin x\cos x \\ \end{gathered}[/tex]Substitute into the expression above
[tex]1+\sin 2x=\frac{25}{16}[/tex]Subtract 1 from both sides
[tex]\begin{gathered} 1-1+\sin 2x=\frac{25}{16}-1 \\ \text{Then} \\ \sin 2x=\frac{9}{16} \end{gathered}[/tex]Hence
[tex]\begin{gathered} \sin 2x=\frac{p}{q}=\frac{9}{16} \\ \\ \text{where } \\ p=9,q=16 \end{gathered}[/tex]Therefore for pq, we have
[tex]pq=p\times q=9\times16=144[/tex]Hence
pq=144
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