Question

1) (sin\(\dfrac{x}{2}\) - cos\(\dfrac{x}{2}\))² + \(\sqrt{3}\)cosx = 2sin5x +1

2) 2sinx(\(\sqrt{3} \)cosx + sinx + 2sin3x)=1

3) (1 + 2cosx)(cosx - \(\sqrt{3}\)sinx) =1

Answer 1

\(\left(sin\dfrac{x}{2}-cox\dfrac{x}{2}\right)^2+\sqrt{3}cosx=2sin5x+1\)

⇔\(sin^2\dfrac{x}{2}+cos^2\dfrac{x}{2}-2sin\dfrac{x}{2}cos\dfrac{x}{2}+\sqrt{3}cosx=2sin5x+1\)

⇔\(1-sinx+\sqrt{3}cosx=2sin5x+1\)

⇔\(sin\left(\dfrac{\Pi}{3}-x\right)=sin5x\)

Answer 2

\(2sinx\left(\sqrt{3}cosx+sinx+2sin3x\right)=1\)

⇔\(2\sqrt{3}sinxcosx+2sin^2x+4sinxsin3x=1\)

⇔\(\sqrt{3}sin2x+1-cos2x+cos2x-2cos4x=1\)

⇔\(\sqrt{3}sin2x+cos2x=2cos4x\)

⇔\(cos\left(2x-\dfrac{\Pi}{3}\right)=cos4x\)

Answer 3

\(\left(1+2cosx\right)\left(cosx-\sqrt{3}sinx\right)=1\)

⇔\(cosx+2cos^2x-\sqrt{3}sinx-2\sqrt{3}sinxcosx=1\)

⇔\(cosx-\sqrt{3}sinx+1+cos2x-\sqrt{3}sin2x=1\)

⇔\(cosx-\sqrt{3}sinx=\sqrt{3}sin2x-cos2x\)

⇔\(sin\left(\dfrac{\Pi}{6}-x\right)=sin\left(2x-\dfrac{\Pi}{6}\right)\)