ĐK $x\neq \frac{\pi}4+k\frac{\pi}2;x\neq \frac{\pi}2+k\pi$$pt\Leftrightarrow \frac{sin3x+sin(x-\frac{\pi}2)}{cos2x}=\frac{\sqrt2(sin^2x+2sinxcosx-cos^2x)}{cos^2x-sin^2x}$$\Leftrightarrow 2sin(2x-\frac{\pi}4)cos(x+\frac{\pi}4)=2sin(2x-\frac{\pi}4)$$\Leftrightarrow x=\frac{\pi}8+k\frac{\pi}2 \veebar x=-\frac{\pi}4+k2\pi$
ĐK $x\neq \frac{\pi}4+k\frac{\pi}2;x\neq \frac{\pi}2+k\pi$$pt\Leftrightarrow \frac{sin3x+sin(x-\frac{\pi}2)}{cos2x}=\frac{\sqrt2(sin^2x+2sinxcosx-cos^2x)}{cos^2x-sin^2x}$$\Leftrightarrow 2sin(2x-\frac{\pi}4)cos(x+\frac{\pi}4)=2sin(2x-\frac{\pi}4)$$\Leftrightarrow x=\frac{\pi}4+k\frac{\pi}2 \veebar x=-\frac{\pi}4+k2\pi$KHĐK => pt vô nghịêm
ĐK $x\neq \frac{\pi}4+k\frac{\pi}2;x\neq \frac{\pi}2+k\pi$$pt\Leftrightarrow \frac{sin3x+sin(x-\frac{\pi}2)}{cos2x}=\frac{\sqrt2(sin^2x+2sinxcosx-cos^2x)}{cos^2x-sin^2x}$$\Leftrightarrow 2sin(2x-\frac{\pi}4)cos(x+\frac{\pi}4)=2sin(2x-\frac{\pi}4)$$\Leftrightarrow x=\frac{\pi}
8+k\frac{\pi}2 \veebar x=-\frac{\pi}4+k2\pi$