Question

Помогите решить уравнение с синусом!

Answer 1

Sin²2x-2sin2xcos2x/2√3-1/2=0
2√3sin²2x-2sin2xcos2x-√3=0
2√3sin²2x-2sin2xcos2x-√3sin²2a-√3cos²2x=0/cos²2x
√3tg²2x-2tg2x-√3=0
tg2x=a
√3a²-2a-√3=0
D=4+12=16
a1=(2-4)/2√3=-1/√3⇒tg2x=-1/√3⇒2x=-π/6+πn⇒x=-π/12+πn/2,n∈z
a2=(2+4)/2√3=√3⇒tg2x=√3⇒2x=π/3+πk⇒x=π/6+πk/2,k∈z
[-π/3;arcctg4/5]=[-π/3;arctg5/4]
x=-π/12
x=π/6