Solve the equation cosec^2(x) = 1 + 2cot(x), for -180° < x ≤ 180°.

First, we want to convert the equation to a simpler form involving just one function of x. Squared trigonometric functions are normally good candidates for simplification, thanks to the trigonometric identities. Using the identity:cot^2(x) + 1 = cosec^2(x),we see the left-hand side simplifies to:cot^2(x) + 1 = 1 + 2cot(x).Rearranging,cot^2(x) – 2cot(x) = 0.We now have a simple quadratic equation in terms of the single variable cot(x). Since there is no constant term, we can factorise the quadratic by taking out a factor of cot(x), leaving:cot(x)*(cot(x) – 2) = 0.By the factor theorem, the solutions are therefore:cot(x) = 0, and cot(x) = 2.We want solutions in the domain -180° < x ≤ 180°. Using that cot(x) = 1/tan(x), we write the above in terms of tan(x) as:tan(x) = ±∞, and tan(x) = 1/2,where we have noted that either a plus or minus infinite value of tan(x) gives cot(x) = 0. These are two different solutions to the equation, which we investigate separately to find all possible solutions in terms of x.1) tan(x) = ±∞: Referring to a graph of tan(x), for example the one at https://commons.wikimedia.org/wiki/File:Tan_proportional.svg, we can easily see the plus and minus infinities are at x = 90° and x = –90°. These are the only solutions in this domain.2) tan(x) = 1/2: The most obvious solution is to take x = arctan(1/2), which we find on a calculator as 26.6°. However looking back at our graph, there is another solution translated by –180°. This second solution is 26.6° – 180° = 153.4°. So, our final answer is x = ±90°, x = 26.6° and x = 153.4°.