Question

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

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

$-1\leq sinx\leq 1\\\\(sinx)_1=\frac{-\sqrt2-3\sqrt{2}}{4}=\frac{-4\sqrt2}{4}=-\sqrt2\approx -1,4<-1\; \to \; \; x_1\in \varnothing \\\\(sinx)_{2}=\frac{-\sqrt2+3\sqrt2}{4}=\frac{2\sqrt2}{4}=\frac{\sqrt2}{2}\; .\\\\x_2=(-1)^{n}\cdot arcsin(\frac{\sqrt2}{2})+\pi n\; ,\; n\in Z\\\\\underline {x=(-1)^{n}\cdot \frac{\pi}{4}+\pi n\; ,\; n\in Z}$

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