Mathematics, 06.07.2019 18:00, jb7692
Solve sin4xcos2x-cos4xsin2x=square root of 2 sinx over the interval [0,2pi)
Answers: 1
![\bf 2sin(x)cos(x)=sin(x)\sqrt{2}\implies 2sin(x)cos(x)-sin(x)\sqrt{2}=0 \\\\\\ sin(x)~[2cos(x)-\sqrt{2}]=0\\\\ -------------------------------\\\\ sin(x)=0\implies \measuredangle x=0~~,~~\pi \\\\ -------------------------------\\\\ 2cos(x)-\sqrt{2}=0\implies 2cos(x)=\sqrt{2}\implies cos(x)=\cfrac{\sqrt{2}}{2} \\\\\\ \measuredangle x=\frac{\pi }{4}~~,~~\frac{7\pi }{4}](/tpl/images/0058/7540/bb3a2.png)
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Solve sin4xcos2x-cos4xsin2x=square root of 2 sinx over the interval [0,2pi)...
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