# \ #
# \ mbox{Before proceeding, I want to say this is a beautiful question.} #
# \ mbox{I confess, when I first saw it, and looked at it briefly, I thought} \ \mbox{it was impossible to solve, meaning no solutions.} #
# \ mbox{Then I looked at it a little more closely, observed that each of} \ \mbox{the trig functions on the LHS of the equation had to equal -1.} \ \mbox{Will elaborate on this shortly. And again, I thought this} \ \mbox{condition was impossible, owing to the phase difference of} \ \mbox{sin and cos.} #
# \mbox{Then I looked at it more deeply.} #
# \mbox{Ok, here we go.} #
# \ #
# \mbox{Given:} \qquad \qquad \qquad \qquad \qquad \quad sin(7x) + cos(2x) = -2. #
# \mbox{1) We know that:} #
# \qquad \qquad \qquad \quad -1 <= sin(x) <= 1 \qquad \mbox{and} -1 <= cos(x) <= 1. #
# \qquad \ \mbox{Hence: } #
# \qquad \qquad \qquad \quad -1 <= sin(7x) <= 1 \qquad \mbox{and} -1 <= cos(2x) <= 1. #
# \qquad \ \mbox{So if:} \qquad \qquad \qquad \qquad \qquad \quad sin(7x) + cos(2x) = -2, #
# \qquad \ \mbox{Then: } #
# \qquad \qquad \qquad \qquad \quad sin(7x) = -1 \qquad \mbox{and} \qquad cos(2x) = -1; #
# \qquad \ \mbox{Otherwise: } #
# \qquad \qquad \qquad \quad -1 < sin(7x) <= 1 \qquad \quad \mbox{or} \qquad -1 < cos(2x) <= 1 #
# \qquad \ \mbox{And so: } #
# \qquad \qquad \qquad \quad (-1) + (-1) < sin(7x) + cos(2x) <= 1 + 1. #
# \qquad \qquad \qquad \qquad \qquad :. \quad -2 < sin(7x) + cos(2x) <= 2. #
# \qquad \ \mbox{In particular, then: } #
# \qquad \qquad \qquad \qquad \qquad :. \quad -2 \ne sin(7x) + cos(2x), #
# \qquad \ \mbox{Which contradicts the given. } #
# \qquad \ \mbox{Thus: } #
# \qquad \qquad \qquad \qquad \quad sin(7x) = -1 \qquad\mbox{and} \qquad cos(2x) = -1. \qquad \qquad (1) #
# \qquad \ \mbox{This will be our starting point ! } #
# \ #
# \mbox{2) So we have now: } #
# \qquad \qquad \qquad \qquad \quad sin(7x) = -1 \qquad\mbox{and} \qquad cos(2x) = -1. #
# \qquad \ \mbox{Thus, by basic facts about sin and cos: } #
# \qquad \qquad \qquad \qquad \quad 7x = {3 \pi}/2 + 2 h \pi, \qquad\mbox{for some integer} \ \ h \mbox{;} \qquad \qquad \mbox{and} #
# \qquad \qquad \qquad \qquad \quad 2x = \pi + 2 k \pi, \qquad \qquad\mbox{for some integer} \ \ k. #
# \qquad \ \mbox{So then:} #
# \qquad \qquad \qquad \qquad \quad x = {3 \pi}/14 + 2/7 h \pi, \qquad\mbox{for some integer} \ \ h \mbox{;} \qquad \qquad \mbox{and} #
# \qquad \qquad \qquad \qquad \quad x = \pi/2 + k \pi, \qquad \qquad \quad \mbox{for some integer} \ \ k. #
# \qquad \ \mbox{So equating these, we have in particular:} #
# \qquad \qquad \qquad \qquad \quad {3 \pi}/14 + 2/7 h \pi \ = \pi/2 + k \pi\qquad\mbox{for some integers} \ \ h, k. #
# \qquad \ \mbox{Dividing by} \ \pi:#
# \qquad \qquad \qquad \qquad \quad 3/14 + 2/7 h \ = 1/2 + k \qquad\mbox{for some integers} \ \ h, k. #
# \qquad \ \mbox{Clearing fractions, and putting the variables all on one side:} #
# \qquad \qquad \qquad \qquad \quad 4 h - 14 k \ = \ 4 \qquad\mbox{for some integers} \ \ h, k. #
# \qquad \ \mbox{Dividing by 2, we have at last:} #
# \qquad \qquad \qquad \qquad \quad 2 h - 7 k \ = \ 2 \qquad\mbox{for some integers} \ \ h, k. \qquad \qquad (2) #
# \qquad \ \mbox{This is the key equation upon which the solution rests.} #
# \qquad \ \mbox{We seek all integer, "nice", solutions of this equation.} #
# \ #
# \qquad \ \mbox{This equation is a linear equation in two variables, where the} \ \mbox{solutions are required to be integers.
This class of equation is} \ \mbox{called a Diophantine equation -- one where integer solutions} \ \mbox{only are desired.} #
# \qquad \ \mbox{A little bit of history. There is a great deal of theory and} \ \mbox{knowledge about these. Some of the deepest reaches of} \ \mbox{mathematics have been developed or found to treat such} \ \mbox{equations. And they have been the subject of revered history,} \ \mbox{going back to the Greeks, and continuing right up until today.} #
# \qquad \ \mbox{Ok. Having said that, the theory of Linear Diophantine} \ \ \mbox{Equations -- such as ours in (2), is well-developed. It does use} \ \mbox{a bit of theory we cannot develop here. But our equation,} \ \mbox{perhaps thankfully, may be simple enough to solve completely,} \ \mbox{without this theory. Onward.} #
# \qquad \ \mbox{I think, for now, I will just give the complete} \ \mbox{solution set for (2) -- holding off on the theory for this} \ \mbox{Diophantine Equation.} #
# \qquad \ \mbox{Ok. Recall equation (2) from above:} #
# \qquad \qquad \qquad \qquad \quad 2 h - 7 k \ = \ 2 \qquad mbox{for some integers} \ \ h, k. \qquad \qquad (2) #
# \qquad \ \mbox{We want all integer solutions of this (all the "nice" solutions).} \ \mbox{The theory of Linear Diophantine Equations gives us the} \ \mbox{complete solution set for (2):} #
# \qquad \qquad \qquad \qquad \qquad \quad h = 1 + 7 t, \qquad k = 2 t; \qquad \qquad t \quad \mbox{any integer.} #
# \qquad \ \mbox{Now we will be finished, when we substitute these values for} \ \ h, k \ \ \mbox{into the solutions for} \ \ x \ \ \mbox{we obtained in the middle of} \ \mbox{part (2) above.} #
# \ #
# \qquad \ \mbox{Let's sum up by recalling these solutions for} \ x \ \mbox{and the} \ \mbox{recent solutions for} \ h, k \ . \mbox{We have:} #
# \qquad \qquad \qquad \qquad \quad x = {3 \pi}/14 + 2/7 h \pi, \qquad\mbox{for some integer} \ \ h \mbox{;} \qquad \qquad \mbox{and} #
# \qquad \qquad \qquad \qquad \quad x = \pi/2 + k \pi, \qquad \qquad \ \ \mbox{for some integer} \ \ k; #
# \qquad \qquad \qquad \qquad \qquad \quad h = 1 + 7 t, \qquad k = 2 t; \qquad \qquad t \quad \mbox{any integer.} #
# \qquad \ \mbox{Putting the solutions for} \quad \h,k \quad \mbox{into the solutions for} \quad x: #
# \qquad \qquad \qquad \qquad \quad x = {3 \pi}/14 + 2/7 ( 1 + 7 t ) \pi, \qquad \qquad \mbox{for some integer} \ \ t \mbox{;} \qquad \qquad \mbox{and} #
# \qquad \qquad \qquad \qquad \quad x = \pi/2 + (2 t) \pi, \qquad \qquad \qquad \qquad \qquad \ \mbox{for some integer} \ \ t. #
# \qquad \ \mbox{Simplifying, we get:}: #
# \qquad \qquad \qquad \qquad \quad x = {3 \pi}/14 + {4\pi}/14 + 2/7( 7 t ) \pi, \qquad\mbox{for some integer} \ \ t \mbox{;} \qquad \qquad \mbox{and} #
# \qquad \qquad \qquad \qquad \quad x = \pi/2 + (2 t) \pi, \qquad \qquad \qquad \qquad \qquad \ \mbox{for some integer} \ \ t. #
# \qquad \ \mbox{Hence:} #
# \qquad \qquad \qquad \qquad \quad x = \pi/2 + 2 \pi t, \qquad\mbox{for some integer} \ \ t \mbox{;} \qquad \qquad \mbox{and} #
# \qquad \qquad \qquad \qquad \quad x = \pi/2 + 2 \pi t, \quad
\ \ \mbox{for some integer} \ \ t. #
# \qquad \ \mbox{And these are the solutions} \ x \ \mbox{of our original equation} \ \mbox{from the start:} #
# \qquad \qquad \qquad \qquad \quad x = \pi/2 + 2 \pi t, \qquad\mbox{for some integer} \ \ t \mbox{;} \qquad \qquad \mbox{and} #
# \qquad \qquad \qquad \qquad \quad x = \pi/2 + 2 \pi t, \quad
\ \ \mbox{for some integer} \ \ t. #
# \qquad \ \mbox{As these formulas (necessarily) come out the same for} \ x \ \mbox{in both equations, we have the one solution for} \ x\ \ \mbox{for our original equation} \ \mbox{from the start:} #
# \qquad \qquad \qquad \qquad \quad x = \pi/2 + 2 \pi t, \quad
\ \ \mbox{for some integer} \ \ t. #
# \ #
# \mbox{Summing up, we finally have:} #
# \mbox{The solution set of:} \qquad \qquad sin(7x) + cos(2x) = -2 #
# \mbox{Is:} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad x = \pi/2 + 2 \pi t, \quad \ \ \mbox{for some integer} \ \ t. #
# \ #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \mbox{QED} \quad \square #
