Question

4sin^3 x +3 sinx+4√3=4√3 cos^2 x
решить,задание из егэ

Answer 1

$4sin^3x+3sinx+4\sqrt{3}=4\sqrt{3}cos^2x\\4sin^3x+3sinx+4\sqrt{3}=4\sqrt{3}(1-sin^2x)\\4sin^3x+3sinx+4\sqrt{3}=4\sqrt{3}-4\sqrt{3}sin^2x\\4sin^3x+4\sqrt{3}sin^2x+3sinx-0\\sinx(4sin^2x+4\sqrt{3}sinx+3)=0\\1)\\sinx=0\\x=\pi n$; n ∈ Z

$2)\\4sin^2x+4\sqrt{3}sinx+3=0\\sinx=t\\4t^2+4\sqrt{3}t+3=0\\(2t+\sqrt{3})^2=0\\t=-\frac{\sqrt{3}}{2}\\sinx=-\frac{\sqrt{3}}{2}\\x=(-1)^n^+^1\frac{\pi }{3}+\pi n$; n ∈ Z