Proof steps please

Find the first step in the following "Proof" that is incorrect, and explain in a short paragraph why the reasoning is incorrect. Statement: The equation 26456+36=64+6236+cos(6) (8) has a positive solution. In other words, there is some positive number 6 satisfying (8}. Proof: By dividing both sides of the Equation (8) by 64 + 62 36 + cos(6) it follows Equation (8) has a solution if 264 562 + 33 64+62 _36+cos(6) 1 (9) does. Let 2 17(9): 26456 +319 (10) 64+6236+cos(6)' From {2.5.4} of the text it follows that cos is a continuous function. By Theorem 2.4.2 (i), (ii) and (iii) it follows that 264 56 + 36 and 64 + 562 + 36 l 005(6) are both continuous functions. By applying Theorem2.4.2 (v) it follows that F is also a continuous function of 6. Then Equation (8} will have a positive solution provided that F(6) : 1 does. This will be established by using Theorem 2.6.1. Begin by noting that if 6 = 1 then 264 562 l 36 = 0 and hence F(1) = 0. While it is a natural idea that that evaluating 264 562 l 36 at 6 = 1 will be useful, finding a value of6 such that 264 562 + 36 > 1 is not that obvious. In such situations it can be useful to evaluate the limit at 6 goes to 00. With this in mind, observe that , 294 592 + 36 _ 6320 64 +62 36+cos(6) 1' 234/64 56/64 + 39/94 1 9330 94/94 + 92/94 _ 39/94 + com/e4 , 2 56-9 + 36/6-3 all{go 1 + 62 363 l (:os(lSl)/64 111119100 2 562 + 36/6-3 I 2. liuzrlg>0 1 l 6'2 36'3 l cos(6)/64 Lete : 1/2. Then there is some K such that |F(6) 2' 3/2 > 1. To summarize, the following facts about F have been shown: . F is continuous at all 6 Z 0 - F(1) : 0 1 such that F((:3) > 1. It then follows from Theorem 2.6.1 that there must be some 6 between 1 and Q such that F(6) = 1. Show that the equation 0/9 + sin(0) /8 = cos-(0/3) + cos (0) + 1 has a positive solution. As usual, you will have to provide the precise reference to the text for the theorems you will use and verify that the hypotheses are satisfied. You should use the correct parts of Question 5 as a guide to your solution