The first part of the question is actually a "show that" the equation in the question can be rearranged to the form 10 cot^(x) + 11 cot(x) - 6 = 0. To do this add 11 cot (x) and subtract 16 to/from both sides of the equation in the question. This gives 10 cosec^2(x) + 11 cot(x) - 16 = 0 which can be written as 10 (cosec^2(x) - 1) + 11 cot(x) - 6 = 0. By substituting in the identity cot^2(x) = cosec^2(x) - 1 you get the required form. If you can remember this identity it can be derived from sin^2(x) + cos^2(x) = 1 as dividing by sin^2(x) gives 1 + cot^2(x) = cosec^2(x).Now for the next part you need to factorise this quadratic equation in the variable cot(x). To do this we want to write the equation in the form (A cot(x) + B)(C cot(x) + D) = 0, where A,B,C,D are numbers. A and C must multiple together to make 10, and B and D similarly to make -6. Also note one of B or D must be negative. Options for A and C are 10 and 1 or 2 and 5. For B and C they are ±6 and ±1 or ±2 and ±3. By considering the other condition that AD + BC = 11 the factorisation must be (5 cot(x) - 2)(2cot(x) + 3) = 0. Then consider when both factors are equal to 0 separately. 5 cot (x) -2 = 0 implies that 5 - 2 tan x) = 0 therefore tan(x) = 5/2. Similarly tan(x) = -2/3.