Ответ:

[tex]2cos^2\alpha+\sqrt3 sin2\alpha-1=2sin \left(\frac{\pi}{6}+2\alpha \right)[/tex]

Объяснение:

[tex]2cos^2\alpha+\sqrt3 sin2\alpha-1=[/tex]

[tex]2cos^2\alpha+\sqrt3 sin2\alpha-(sin^2\alpha+cos^2\alpha)=[/tex]

[tex]2cos^2\alpha+\sqrt3 sin2\alpha-sin^2\alpha-cos^2\alpha=[/tex]

[tex]cos^2\alpha+\sqrt3 sin2\alpha-sin^2\alpha=[/tex]

[tex]cos2\alpha+2\cdot\frac{\sqrt3}{2}sin 2\alpha[/tex]

[tex]cos2\alpha+2sin\frac{\pi}{3}sin 2\alpha=[/tex]

[tex]2\cdot \frac{1}{2}cos2\alpha+2cos\frac{\pi}{6}sin 2\alpha=[/tex]

[tex]2\cdot sin\frac{\pi}{6}cos2\alpha+2cos\frac{\pi}{6}sin 2\alpha=[/tex]

[tex]2\cdot  \left(sin\frac{\pi}{6}cos2\alpha+cos\frac{\pi}{6}sin 2\alpha\right) =[/tex]

[tex]2sin \left(\frac{\pi}{6}+2\alpha \right) [/tex]