Question

Помогите решить 3 примера!


cos(pi/9-4x)=1


sinx/7=0


sinx-√3cosx=1

Answer 1

$1.\cos(frac{pi}{9}-4x)=1\4x-frac{pi}{9}=2pi n\4x=frac{pi}{9}+2pi n\x=frac{pi}{36}+frac{pi n}{2}, ; nin Z.$

$2.\sinfrac{x}{7}=0\frac{x}{7}=pi n\x=7 pi n, ; nin Z;\\ 3.\sinx-sqrt3cosx=1| ; :2\frac{1}{2}sinx-frac{sqrt3}{2}cosx=frac{1}{2}\frac{1}{2}=cosfrac{pi}{3}, quad frac{sqrt3}{2}=sinfrac{pi}{3};\sinxcosfrac{pi}{3}+cosxsinfrac{pi}{3}=frac{1}{2}\sin(x+frac{pi}{3})=frac{1}{2}\x+frac{pi}{3}=(-1)^{n}arcsinfrac{1}{2}+pi n\x=(-1)^nfrac{pi}{6}-frac{pi}{3}+pi n, ; nin Z;$

$x+frac{pi}{3}=frac{pi}{6}+2pi n\x=frac{pi}{2}+2pi n, ; nin Z;\\x+frac{pi}{3}=pi-frac{pi}{6}+2pi n\x+frac{pi}{3}=frac{5pi}{6}+2pi n\x=frac{pi}{2}+2pi n, ; nin Z.$