Value of $\,x\,,\,$ for $\,\cos x>0\,.$
In my book the value of $\,x\,$ for $\,\cos x>0\,$ is given as :
$$\bigcup_{n=-\infty}^\infty \left(\left(2n-\frac12\right)\pi, \left(2n+1\right)\frac\pi2\right)$$
How did they get the domain for this ?
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$\begingroup$The $\cos(x)+|\cos(x)|$ in the denominator and square root imply that we must have $$\cos(x)+|\cos(x)|>0.$$ Otherwise, we would be dividing by $0$ or taking the square root of a negative number.
Now, we just need to find when this inequality holds. If $\cos(x)\leq 0$, then $|\cos(x)|$ is in fact equal to $-\cos(x)$, so the inequality does not hold.
Therefore we simply need $\cos(x)>0$. The solution to this inequality can be found by graphing $y=\cos(x)$, or by using the unit circle, or whatever method of understanding the cosine function you are most comfortable with.
EDIT based on comment*
To find the solution to $\cos(x)>0$, remember that within the unit circle, $\cos(\theta)$ is represented by the $x$ coordinate. So we are looking for the angles that point to the right, which are $(-\pi/2,\pi/2)$. Then, you can add or subtract any multiple of $2\pi$ to an angle to get the same cosine value, so for any $n$, we have $(2\pi n -\pi/2,2\pi n + \pi/2)$ also satisfies $\cos(x)>0$. The domain is the union of all of these intervals for any $-\infty < n< \infty.$ @ThomasAndrews correctly pointed out a typo in the domain you have shown. It should be $$\bigcup_{n=-\infty}^{\infty} \bigg(\big(2n-1/2\big)\pi,\big(2n+1/2\big)\pi\bigg).$$
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