+564 On the interval $$[0,2\pi]$$, where is the function
$$f(x)=\sqrt{2sinx-\sqrt{3}}$$ continuous?
+118687 We are talking real numbers here so the number under the square root must be positive.
$$\\2sinx-\sqrt3\ge0\\\\
2sinx\ge\sqrt3\\\\
sinx\ge\frac{\sqrt3}{2}\\\\$$
for starters sinx must be positive so it must be 1st and 2nd quad only.
$$asin(\frac{\sqrt3}{2})=\frac{\pi}{3}$$ the second quadrant equivalent is $$\pi-\frac{\pi}{3}=\frac{2\pi}{3}$$
$$\frac{\pi}{3}\le x \le \frac{2\pi}{3}$$
I know this because sin(pi/2)=1 and I know what the sine curve looks like so I know the answer will be between these two values. 
+118687 We are talking real numbers here so the number under the square root must be positive.
$$\\2sinx-\sqrt3\ge0\\\\
2sinx\ge\sqrt3\\\\
sinx\ge\frac{\sqrt3}{2}\\\\$$
for starters sinx must be positive so it must be 1st and 2nd quad only.
$$asin(\frac{\sqrt3}{2})=\frac{\pi}{3}$$ the second quadrant equivalent is $$\pi-\frac{\pi}{3}=\frac{2\pi}{3}$$
$$\frac{\pi}{3}\le x \le \frac{2\pi}{3}$$
I know this because sin(pi/2)=1 and I know what the sine curve looks like so I know the answer will be between these two values.
