e: \(\sin x\cdot cosx+\sqrt3\cdot\sin2x=0\)
=>\(\frac12\cdot\sin2x+\sqrt3\cdot\sin2x=0\)
=>\(\sin2x\left(\sqrt3+\frac12\right)=0\)
=>sin 2x=0
=>\(2x=k\pi\)
=>\(x=\frac{k\pi}{2}\)
x: Đặt \(a=\sin x-cosx\)
=>\(a^2=\left(\sin x-cosx\right)^2=\sin^2x+cos^2x-2\cdot\sin x\cdot cosx=1-\sin2x\)
=>\(\sin2x=1-a^2\)
\(\sin2x+2\sqrt2\left(\sin x-cosx\right)-3=0\)
=>\(1-a^2+2\sqrt2\cdot a-3=0\)
=>\(-a^2+2\sqrt2a-2=0\)
=>\(a^2-2\sqrt2\cdot a+2=0\)
=>\(\left(a-\sqrt2\right)^2=0\)
=>\(a-\sqrt2=0\)
=>\(a=\sqrt2\)
=>\(\sin x-cosx=\sqrt2\)
=>\(\sqrt2\cdot\sin\left(x-\frac{\pi}{4}\right)=\sqrt2\)
=>\(\sin\left(x-\frac{\pi}{4}\right)=1\)
=>\(x-\frac{\pi}{4}=\frac{\pi}{2}+k2\pi\)
=>\(x=\frac34\pi+k2\pi\)
