Question

sqrt(2)*sin(x)+2*sin(2x-pi/6)=sqrt(3)*sin(2*x)+1
Помогите пожалуйста

Answer 1

$sqrt2sinx+2, sin(2x-frac{pi}{6})=sqrt3, sin2x+1\\sqrt2, sinx+2(sin2x, cosfrac{pi}{6}-cos2x, sinfrac{pi}{6})=sqrt3, sin2x+1\\sqrt2sinx+sqrt3, sin2x-cos2x=sqrt3, sin2x+1\\sqrt2, sinx-(1+cos2x)=0\\sqrt2, sinx-(underbrace {sin^2x+cos^2x}_{1}+underbrace {cos^2x-sin^2x}_{cos2x})=0\\sqrt2, sinx-2cos^2x=0\\sqrt2, sinx-2(1-sin^2x)=0\\2sin^2x+sqrt2, sinx-2=0; ; ,; ; D=2+4cdot 2cdot 2=18; ,; ; sqrt{18}=sqrt{9cdot 2}=3sqrt2$

$-1leq sinxleq 1\\(sinx)_1=frac{-sqrt2-3sqrt{2}}{4}=frac{-4sqrt2}{4}=-sqrt2approx -1,4<-1; to ; ; x_1in varnothing \\(sinx)_{2}=frac{-sqrt2+3sqrt2}{4}=frac{2sqrt2}{4}=frac{sqrt2}{2}; .\\x_2=(-1)^{n}cdot arcsin(frac{sqrt2}{2})+pi n; ,; nin Z\\underline {x=(-1)^{n}cdot frac{pi}{4}+pi n; ,; nin Z}$

$P.S.; quad ; cosfrac{pi}{6}=frac{sqrt3}{2}; ; ,; ; sin frac{pi}{6}=frac{1}{2}$

Answer 2

формулы надо знать