Question

Помогите решить буквы в) д) е)

Answer 1

$1)quadsqrt2cosleft(2x-frac{pi}5right)-1=0Rightarrowsqrt2cosleft(2x-frac{pi}5right)=1Rightarrow\Rightarrowcosleft(2x-frac{pi}5right)=frac1{sqrt2}=frac{sqrt2}2Rightarrow2x-frac{pi}5=frac{pi}4+2pi nRightarrow\Rightarrow2x=frac{pi}4+frac{pi}5+2pi n=frac{9pi}{20}+2pi nRightarrow x=frac{9pi}{40}+pi n,quad ninmathbb{Z}$

$2)quad3+sqrt3tgleft(frac{pi}3-frac x4right)=0Rightarrowsqrt3tgleft(frac{pi}3-frac x4right)=-3Rightarrow\Rightarrow tgleft(frac{pi}3-frac x4right)=-frac3{sqrt3}=-sqrt3Rightarrowfrac{pi}3-frac x4right=-frac{pi}3+pi nRightarrow\Rightarrowfrac x4=frac{pi}3+frac{pi}3+pi n=frac{2pi}3+pi nRightarrow x=frac{8pi}3+3pi n,quadninmathbb{Z}$

$3)quad2cos^25x-sin5x-1=0\cos^25x=1-sin^25xRightarrow\Rightarrow2cos^25x-sin5x-1=2-2sin^25x-sin5x-1=\=-2sin^25x-sin5x+1=0\2sin^25x+sin5x-1=0\sin5x=t\2t^2+t-1=0\D=1+4cdot2=9\t_1=frac{-1+3}{4}=frac12\t_2=frac{-1-3}4=-1\sin5x=frac12Rightarrow5x=frac{pi}6+2pi nRightarrow x=frac{pi}{30}+frac25n,quad ninmathbb{Z}\sin5x=-1Rightarrow5x=frac{3pi}2+2pi nRightarrowx=frac{3pi}{10}+frac25n,quad ninmathbb{Z}$