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Решите задачу:

$\sqrt{108}cos^2 \frac{13 \pi }{12}- \sqrt{27}=\sqrt{108}cos^2( \pi+ \frac{\pi}{12})}- \sqrt{27}= \\ \sqrt{108}cos^2(-\frac{\pi}{12})}- \sqrt{27}=\sqrt{108}* \frac{1+cos\frac{\pi}{6}}{2}- \sqrt{27}= \\ \sqrt{108}* \frac{1+\frac{ \sqrt{3}}{2}}{2}- \sqrt{27}=\sqrt{4*27}* \frac{2+ \sqrt{3}}{4}}- \sqrt{27}= \\ \sqrt{27}* \frac{2+ \sqrt{3}}{2}}- \sqrt{27}=\sqrt{27}( \frac{2+ \sqrt{3}}{2}}-1)=\sqrt{27} \frac{2+ \sqrt{3}-2}{2}= \frac{ \sqrt{27}* \sqrt{3}}{2}= \\ \frac{ \sqrt{3*3^3}}{2}= \frac{9}{2}=4.5$