

[tex]\sqrt{2}(sinx+\sqrt{3}cosx)=\sqrt{3}cos2x-sin2x[/tex]
[tex]\Leftrightarrow \sqrt{2}cos(x-\frac{\pi}{6})=-sin(2x-\frac{\pi}{3})\\\Leftrightarrow \sqrt{2}cos(x-\frac{\pi}{6})=-2sin(x-\frac{\pi}{6})cos(x-\frac{\pi}{6})\\\Leftrightarrow ...[/tex]
[tex]\Leftrightarrow \sqrt{2}cos(x-\frac{\pi}{6})=-sin(2x-\frac{\pi}{3})\\\Leftrightarrow \sqrt{2}cos(x-\frac{\pi}{6})=-2sin(x-\frac{\pi}{6})cos(x-\frac{\pi}{6})\\\Leftrightarrow ...[/tex]
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