Помогите решить буквы в) д) е)

Решите задачу:

$1)\quad\sqrt2\cos\left(2x-\frac{\pi}5\right)-1=0\Rightarrow\sqrt2\cos\left(2x-\frac{\pi}5\right)=1\Rightarrow\\\Rightarrow\cos\left(2x-\frac{\pi}5\right)=\frac1{\sqrt2}=\frac{\sqrt2}2\Rightarrow2x-\frac{\pi}5=\frac{\pi}4+2\pi n\Rightarrow\\\Rightarrow2x=\frac{\pi}4+\frac{\pi}5+2\pi n=\frac{9\pi}{20}+2\pi n\Rightarrow x=\frac{9\pi}{40}+\pi n,\quad n\in\mathbb{Z}$

$2)\quad3+\sqrt3tg\left(\frac{\pi}3-\frac x4\right)=0\Rightarrow\sqrt3tg\left(\frac{\pi}3-\frac x4\right)=-3\Rightarrow\\\Rightarrow tg\left(\frac{\pi}3-\frac x4\right)=-\frac3{\sqrt3}=-\sqrt3\Rightarrow\frac{\pi}3-\frac x4\right=-\frac{\pi}3+\pi n\Rightarrow\\\Rightarrow\frac x4=\frac{\pi}3+\frac{\pi}3+\pi n=\frac{2\pi}3+\pi n\Rightarrow x=\frac{8\pi}3+3\pi n,\quadn\in\mathbb{Z}$

$3)\quad2\cos^25x-\sin5x-1=0\\\cos^25x=1-\sin^25x\Rightarrow\\\Rightarrow2\cos^25x-\sin5x-1=2-2\sin^25x-\sin5x-1=\\=-2\sin^25x-\sin5x+1=0\\2\sin^25x+\sin5x-1=0\\\sin5x=t\\2t^2+t-1=0\\D=1+4\cdot2=9\\t_1=\frac{-1+3}{4}=\frac12\\t_2=\frac{-1-3}4=-1\\\sin5x=\frac12\Rightarrow5x=\frac{\pi}6+2\pi n\Rightarrow x=\frac{\pi}{30}+\frac25n,\quad n\in\mathbb{Z}\\\sin5x=-1\Rightarrow5x=\frac{3\pi}2+2\pi n\Rightarrowx=\frac{3\pi}{10}+\frac25n,\quad n\in\mathbb{Z}$