1. ## Finding the domain

The question:
Find the domain of:
$\displaystyle \sqrt{1 - 2sinx}$

My attempt:
$\displaystyle 1 - 2sinx = 0$
$\displaystyle -2sinx = -1$
$\displaystyle 2sinx = 1$
$\displaystyle sinx = \frac{1}{2}$

I worked out that x = 30 degrees, or Pi/6

So I want the values where sinx is larger than or equal to 1/2:
$\displaystyle \frac{\pi}{6} + 2\pi k <= x <= \frac{5\pi}{6} + 2\pi k$
Where k is an integer.

I thought this made sense, but the answer in my book is:
$\displaystyle -\frac{7\pi}{6} + 2\pi k <= x <= \frac{\pi}{6} + 2\pi k$

What am I doing wrong? :/

2. Hello, Glitch!

The book's answer is wrong . . .

Find the domain of: .$\displaystyle f(x) \:=\:\sqrt{1 - 2\sin x}$

Use inequalities . . .

. . $\displaystyle 1 - 2\sin x \:\ge \:0 \quad\Rightarrow\quad \text{-}2\sin x \:\ge \:\text{-}1 \quad\Rightarrow\quad \sin x \:\le \frac{1}{2}$

If $\displaystyle \sin x$ is between $\displaystyle \text{-}\frac{1}{2}$ and $\displaystyle \frac{1}{2}$

. . then $\displaystyle x$ is basically between $\displaystyle \text{-}\frac{\pi}{6}$ and $\displaystyle \frac{\pi}{6}$

Therefore: .$\displaystyle \text{-}\frac{\pi}{6} + 2\pi k \;\leq \;x \;\leq\; \frac{\pi}{6} + 2\pi k$

3. How did you get the -1/2?

4. Hi,
Glitch, Your solution is where the sine is greater than or equal to 1/2! Infact you should find the angles which their sine is less than or equal to 1/2.
The simplest way is to use a unit circle or a sine graph to solve this Trigonometric inequality.
1-Draw a unit circle.
2-Draw a horizontal line to cosine axis such that intersects the sine axis at the point 1/2.
3-Since at the 1st and 2nd quadrants sine is 1/2 at pi/6 and pi-pi/6=5pi/6, the line intersects the circle at those points. Now you can see that all the angles between pi/6 and 5pi/6 have sines greater than 1/2. So what interval do you think you should takeas solution?
Therefore you can see that the book's solution is true! (Note that -7pi/6 is in the same location of 5pi/6 and so they have the equal Trig relationships)

5. And Soroban, you should note that when sine<=1/2 then it could has all the values in [-1,1/2]. Infact you forgot the numbers in [-1,-1/2).

6. Thank you very much! As you say, I was looking at when it was larger than 1/2 like a fool. I got the answer the textbook states.

My method was to draw the sine graph and get an idea of what was happening. Hopefully that's a satisfactory method.