solve the equation for all exact solutions where appropriate. round approximate answers in degrees to the…

solve the equation for all exact solutions where appropriate. round approximate answers in degrees to the nearest tenth. write answers using the least possible nonnegative angle measures.\n\n2 sin x -√3 = 0\n\nchoose the correct answer below.\n\na. {2π/3 +nπ, where n is any integer}\n\nb. {π/3 +nπ, 2π/3 +nπ, where n is any integer}\n\nc. {π/3 +2nπ, 2π/3 +2nπ, where n is any integer}\n\nd. {π/3 +2nπ, where n is any integer}

solve the equation for all exact solutions where appropriate. round approximate answers in degrees to the nearest tenth. write answers using the least possible nonnegative angle measures.\n\n2 sin x -√3 = 0\n\nchoose the correct answer below.\n\na. {2π/3 +nπ, where n is any integer}\n\nb. {π/3 +nπ, 2π/3 +nπ, where n is any integer}\n\nc. {π/3 +2nπ, 2π/3 +2nπ, where n is any integer}\n\nd. {π/3 +2nπ, where n is any integer}

Answer

Explanation:

Step1: Solve for (\sin x)

Given (2\sin x-\sqrt{3} = 0), add (\sqrt{3}) to both sides: (2\sin x=\sqrt{3}). Then divide both sides by (2): (\sin x=\frac{\sqrt{3}}{2}).

Step2: Find the general solutions

We know that (\sin x=\frac{\sqrt{3}}{2}) when (x = \frac{\pi}{3}+ 2n\pi) (in the first - quadrant) and (x=\frac{2\pi}{3}+2n\pi) (in the second - quadrant) for (n\in\mathbb{Z}) (because the sine function has a period of (2\pi)).

Answer:

C. (\left{\frac{\pi}{3}+2n\pi,\frac{2\pi}{3}+2n\pi,\text{ where }n\text{ is any integer}\right})