Question

Solve sin4xcos2x-cos4xsin2x=square root of 2 sinx over the interval [0,2pi)

Answer 1

$\bf sin({{ \alpha}} - {{ \beta}})=sin({{ \alpha}})cos({{ \beta}})- cos({{ \alpha}})sin({{ \beta}}) \\\\\\ sin(2\theta)=2sin(\theta)cos(\theta)\\\\ -------------------------------\\\\ sin(4x)cos(2x)-cos(4x)sin(2x)=\sqrt{2}sin(x) \\\\\\ sin(4x~~-~~2x)=sin(x)\sqrt{2}\implies sin(2x)=sin(x)\sqrt{2}$

$\bf 2sin(x)cos(x)=sin(x)\sqrt{2}\implies 2sin(x)cos(x)-sin(x)\sqrt{2}=0 \\\\\\ sin(x)~[2cos(x)-\sqrt{2}]=0\\\\ -------------------------------\\\\ sin(x)=0\implies \measuredangle x=0~~,~~\pi \\\\ -------------------------------\\\\ 2cos(x)-\sqrt{2}=0\implies 2cos(x)=\sqrt{2}\implies cos(x)=\cfrac{\sqrt{2}}{2} \\\\\\ \measuredangle x=\frac{\pi }{4}~~,~~\frac{7\pi }{4}$

now, we're not including the III and II quadrants, where the cosine has an angle of the same value, but is negative, because the exercise seems to be excluding the negative values of √(2).