Find all solutions of sin x - square 3-3 sin^2x=0

Answer:
C) [tex]60^\circ+n360^\circ,120^\circ+n360^\circ[/tex]
Step-by-step explanation:
[tex]sinx-\sqrt{3-3sin^2x}=0[/tex]
[tex]sinx=\sqrt{3-3sin^2x} [/tex]
[tex]sin^2x=3-3sin^2x[/tex]
[tex]4sin^2x=3[/tex]
[tex]sin^2x=\frac{3}{4} [/tex]
[tex]sinx=\pm\sqrt{\frac{3}{4} } [/tex]
[tex]sinx=\pm\frac{\sqrt{3}}{2} [/tex]
[tex]x=\frac{\pi}{3}+2\pi n,\frac{2\pi}{3}+2\pi n [/tex]
[tex]x=60^\circ+n360^\circ,120^\circ+n360^\circ[/tex]
Therefore, the correct option is C