Question

[Nguyễn Anh Tuấn]

Giải giup em bai nay voi

Answer 1

[Nguyễn Anh Tuấn]

câu 2a đó ạ

Answer 2

[Sonboygaming Tran]

dùng ông thức hạ bậc

cos²a=\(\dfrac{1+cos2a}{2}\)

pt<=>1+cos(4x+\(\dfrac{2\Pi}{3}\))-3sin(2x+\(\dfrac{5\Pi}{6}\))+1=0

<=>-\(\dfrac{1}{2}\)cos4x-\(\dfrac{\sqrt{3}}{2}\)sin4x+\(\dfrac{3\sqrt{3}}{2}\)sin2x-\(\dfrac{3}{2}\)cos2x+2=0

<=>(-\(\dfrac{1}{2}\)cos4x+\(\dfrac{3\sqrt{3}}{2}\)sin2x+2)+(-\(\sqrt{3}\)sin2x.cos2x-\(\dfrac{3}{2}\)cos2x)=0

<=>[-\(\dfrac{1}{2}\)(1-2sin²2x)+\(\dfrac{3\sqrt{3}}{2}\)sin2x+2)-cos2x.(\(\sqrt{3}\)sin2x+\(\dfrac{3}{2}\))=0

<=>(sin²2x+\(\dfrac{3\sqrt{3}}{2}\)sin2x+\(\dfrac{3}{2}\))-cos2x.(\(\sqrt{3}\)sin2x+\(\dfrac{3}{2}\))=0

<=>(sin2x+\(\dfrac{\sqrt{3}}{2}\))(sin2x+\(\sqrt{3}\))-cos2x.(sin2x+\(\dfrac{\sqrt{3}}{2}\))=0

<=>(sin2x+\(\dfrac{\sqrt{3}}{2}\))(sin2x-cos2x+\(\sqrt{3}\))=0

tới đây bạn tự giải nhé