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

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

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

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

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

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

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

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

⇔\(\sqrt{3}sin2x+cos2x=2cos4x\)

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

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

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

⇔\(1-sinx+\sqrt{3}cosx=2sin5x+1\)

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

Ran Mori trả lời gần đúng

Chả liên quan

tan3x.tanx = 1

⇔tan3x = cotx

⇔\(tan3x=tan\left(\dfrac{\Pi}{2}-x\right)\)

b.\(\left(2cosx-1\right)cotx=\dfrac{3}{sinx}+\dfrac{2sinx}{cosx-1}\left(1\right)\)

ĐKXĐ: sinx≠0 và cosx≠1

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

⇔cosx(2cosx-1)(cosx-1) = 3(cosx-1) + 2sin²x

⇔2cos³x - cos²x - 2cosx +1 = 0

⇔ (cosx-1)(cosx+1)(2cosx-1)=0

chỉ có 1 chữ C vì C này in hoa