Question

4cos^3x+ 3sin(x-пи/2)= 0 на промежутке от -2пи до -пи

Answer 1

4cos³x-3cosx=0
cosx(4cos²x-3)=0
cosx=0⇒x=π/2+πn
(2cosx-√3)(2+cosx+√3)=0
cosx=√3/2⇒x=+-π/6+2πn
cosx=-√3/2⇒x=+-5π/6+2πn