Question

sin^2x-14=0 помогите решить

Answer 1

$sin^2x-dfrac{1}{4}=0; ~~~(sin x-dfrac{1}{2})(sin x + dfrac{1}{2})=0$

$sin~x=pm dfrac{1}{2}\ \ x_1=pm dfrac{pi}{6} +2pi n,~~~x_2=pmdfrac{5pi}{6} +2pi k;~~n,kin Z$

Answer 2

Применим формулу понижения степеней

$displaystyle tt frac{1-cos2x}{2} -frac{1}{4}=0~~bigg|cdot 4\ \ 2(1-cos 2x)-1=0\ \ 2-2cos 2x-1=0\ \ -2cos 2x+1=0\ \ cos 2x=0.5\ \ 2x=pmfrac{pi}{3}+2pi n,n in mathbb{Z}\ \ boxed{tt x=pmfrac{pi}{6}+pi n,n in mathbb{Z}}$