[tex](9\sin2x+9\cos2x)^2=81[/tex]Taking the square root of both sides gives two possible cases,[tex]9\sin2x+9\cos2x=9\implies\sin2x+\cos2x=1[/tex]or[tex]9\sin2x+9\cos2x=-9\implies\sin2x+\cos2x=-1[/tex]Recall that[tex]\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta[/tex]If [tex]\alpha=2x[/tex] and [tex]\beta=\dfrac\pi4[/tex], we have[tex]\sin\left(2x+\dfrac\pi4\right)=\dfrac{\sin2x+\cos2x}{\sqrt2}[/tex]so in the equations above, we can write[tex]\sin2x+\cos2x=\sqrt2\sin\left(2x+\dfrac\pi4\right)=\pm1[/tex]Then in the first case,[tex]\sqrt2\sin\left(2x+\dfrac\pi4\right)=1\implies\sin\left(2x+\dfrac\pi4\right)=\dfrac1{\sqrt2}[/tex][tex]\implies2x+\dfrac\pi4=\dfrac\pi4+2n\pi\text{ or }\dfrac{3\pi}4+2n\pi[/tex](where [tex]n[/tex] is any integer)[tex]\implies2x=2n\pi\text{ or }\dfrac\pi2+2n\pi[/tex][tex]\implies x=n\pi\text{ or }\dfrac\pi4+n\pi[/tex]and in the second,[tex]\sqrt2\sin\left(2x+\dfrac\pi4\right)=-1\implies\sin\left(2x+\dfrac\pi4\right)=-\dfrac1{\sqrt2}[/tex][tex]\implies2x+\dfrac\pi4=-\dfrac\pi4+2n\pi\text{ or }-\dfrac{3\pi}4+2n\pi[/tex][tex]\implies2x=-\dfrac\pi2+2n\pi\text{ or }-\pi+2n\pi[/tex][tex]\implies x=-\dfrac\pi4+n\pi\text{ or }-\dfrac\pi2+n\pi[/tex]Then the solutions that fall in the interval [tex][0,2\pi)[/tex] are[tex]x=0,\dfrac\pi4,\dfrac\pi2,\dfrac{3\pi}4,\pi,\dfrac{5\pi}4,\dfrac{3\pi}2,\dfrac{7\pi}4[/tex]