Trigonometric Systems of Equations

477

Chapter 6: Trigonometric Systems of Equations Section 1: Formulas, Facts, Results The variety of solvable trigonometric systems of equations is too vast to include in a single math book. Our goal in this Chapter is to tackle certain families of systems that at least some of which, many a student and teacher come across in the course of their mathematical lives. Broadly speaking one can divide trigonometric systems into two large categories, even though other categories do exist: albeit, they would be of a marginal interest to a wider mathematical audience. In the first category, the systems found are of the form ⎧f (g1 (x ), g 2 (y )) = 0⎫ ⎨ ⎬ ; where f (t1 , t 2 ) is an algebraic function in two real-valued ⎩ h (x , y ) = 0 ⎭

variables t1 and t 2 (typically a polynomial function; but sometimes it may involve radicals); g1 (x ) and g 2 (y ) are among the six fundamental trigonometric functions; and h(x,y) an algebraic function of two variables. A solution to a trigonometric system, is an ordered pair (xo, yo) of real numbers, which satisfies the given system. The solution set S, a subset of R x R, is the set of all solutions to the system at hand. ⎧f (g1 (x ), g 2 (y )) = 0 ⎫ ⎬; ( ( ) ( ) ) = h x , y 0 g g 4 ⎩ 3 ⎭

In the second category, we find systems of the form ⎨

g1 , g 2 , g3 , g 4 being among the six fundamental trigonometric functions. In Part 1,

we take a look at certain first category systems; and we thoroughly analyze a few of them. In Part 2, we present some second category systems and we determine

478

the solution sets of six of them. More first and second category systems are worked out as solved problems in Section 3; and a good number of them are listed as unsolved problems in Section 4. PART 1: First Category Systems In this part, we present and study six groups of trigonometric systems. We list these groups. Group 1 ⎧sin x + sin y = α ⎫ ⎧sin x − sin y = α ⎫ ⎬ (ii) ⎨ ⎬ ⎩x + y = β ⎭ ⎩x + y = β ⎭

(i) ⎨

⎧sin x + sin y = α ⎫ ⎬ ⎩x − y = β ⎭

(iii) ⎨

⎧sin x − sin y = α ⎫ ⎧cos x + cos β = α ⎫ ⎬ (v) ⎨ ⎬ ⎩x − y = β ⎭ ⎩x + y = β ⎭

⎧cos x − cos y = α ⎫ ⎬ ⎩x + y = β ⎭

(iv) ⎨

⎧cos x + cos y = α ⎫ ⎬ ⎩x − y = β ⎭

(vii) ⎨

(vi) ⎨

⎧cos x − cos y = α ⎫ ⎬ ⎩x − y = β ⎭

(viii) ⎨

Group 2 ⎧sin x sin y = α ⎫ ⎧sin x sin y = α ⎫ ⎬ (ii) ⎨ ⎬ ⎩x + y = β ⎭ ⎩x − y = β ⎭

(i) ⎨

⎧sin x cos y = α ⎫ ⎬ ⎩x + y = β ⎭

(iii) ⎨

⎧sin x cos y = α ⎫ ⎧cos x cos y = α ⎫ ⎬ (v) ⎨ ⎬ ⎩x − y = β ⎭ ⎩x + y = β ⎭

(iv) ⎨

⎧cos x cos y = α ⎫ ⎬ ⎩x − y = β ⎭

(vi) ⎨

Group 3 ⎧ sin x ⎫ ⎧ sin x ⎫ = α⎪ = α⎪ ⎪ ⎪ (i) ⎨ sin y ⎬ (ii) ⎨ sin y ⎬ ⎪x + y = β ⎪ ⎪x − y = β ⎪ ⎩ ⎭ ⎩ ⎭

⎧ cos x ⎫ = α⎪ ⎪ (iii) ⎨ cos y ⎬ ⎪x + y = β ⎪ ⎩ ⎭

⎧ cos x ⎫ ⎧ sin x sin y ⎫ = α⎪ = α⎪ ⎪ ⎪ (iv) ⎨ cos y ⎬ (v) ⎨ cos x cos y ⎬ ⎪x − y = β ⎪ ⎪x + y = β ⎪ ⎩ ⎭ ⎩ ⎭

⎧ sin x sin y ⎫ = α⎪ ⎪ (vi) ⎨ cos x cos y ⎬ ⎪x − y = β ⎪ ⎩ ⎭

479 ⎧ sin x cos y ⎫ = α⎪ ⎪ (vii) ⎨ cos x sin y ⎬ ⎪x + y = β ⎪ ⎩ ⎭

⎧ sin x cos y ⎫ = α⎪ ⎪ (viii) ⎨ cos x sin y ⎬ ⎪x − y = β ⎪ ⎩ ⎭

Group 4 ⎧tan x + tan y = α ⎫ ⎧tan x − tan y = α ⎫ ⎬ (ii) ⎨ ⎬ ⎩x + y = β ⎭ ⎩x − y = β ⎭

(i) ⎨

⎧cot x + cot y = α ⎫ ⎬ ⎩x + y = β ⎭

(iii) ⎨

⎧cot x − cot y = α ⎫ ⎬ ⎩x − y = β ⎭

(iv) ⎨

Group 5 ⎧tan x + tan y = α ⎫ ⎧tan x − tan y = α ⎫ ⎬ (ii) ⎨ ⎬ ⎩x − y = β ⎭ ⎩x + y = β ⎭

(i) ⎨

⎧cot x + cot y = α ⎫ ⎬ ⎩x − y = β ⎭

(iii) ⎨

⎧cot x − cot y = α ⎫ ⎬ ⎩x + y = β ⎭

(iv) ⎨

Group 6 ⎧ tan x ⎫ ⎧ tan x ⎫ = α⎪ = α⎪ ⎪ ⎪ (i) ⎨ tan y ⎬ (ii) ⎨ tan y ⎬ ⎪x + y = β ⎪ ⎪x − y = β ⎪ ⎩ ⎭ ⎩ ⎭

⎧ cot x ⎫ = α⎪ ⎪ (iii) ⎨ cot y ⎬ ⎪x + y = β ⎪ ⎩ ⎭

⎧ cot x ⎫ = α⎪ ⎧tan x tan y = α ⎫ ⎪ (iv) ⎨ cot y ⎬ (v) ⎨ ⎬ ⎩x + y = β ⎭ ⎪x − y = β ⎪ ⎩ ⎭ ⎧cot x cot y = α ⎫ ⎬ ⎩x + y = β ⎭

(vii) ⎨

⎧tan x tan y = α ⎫ ⎬ ⎩x − y = β ⎭

(vi) ⎨

⎧cot x cot y = α ⎫ ⎬ ⎩x − y = β ⎭

(viii) ⎨

480

Group 1 : Analysis and Determination of Solution Sets As it will become evident, it is only necessary to solve the first two systems in this group; (i) and (ii) that is. The other six systems reduce to these two; so their solution sets are readily determined from the solution sets of the first two. The same phenomenon occurs in the other groups; all systems in each group, reduce to one or two systems in those groups. i) Solution of System (i) ⎧ ⎫ ⎛x + y⎞ ⎛x −y⎞ ⎟ cos⎜ ⎟ = α⎪ ⎧sin x + sin y = α ⎫ ⎪2 sin ⎜ We have, ⎨ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎬⇔⎨ ⎬⇔ ⎩x + y = β ⎭ ⎪x + y = β ⎪ ⎩ ⎭

⎧ ⎛ β ⎞ ⎛ x − y ⎞⎟ = α ⎫ ⎪2 sin ⎜ 2 ⎟ cos⎜ ⎪ ⎝ ⎠ ⎝ 2 ⎠ ⇔⎨ ⎬ ⎪x + y = β ⎪ ⎩ ⎭

(1)

Case 1: sin ⎛⎜ β 2 ⎞⎟ = 0 ⇔ β 2 = ρπ ; β = 2ρπ , for some ρεZ . ⎝ ⎠ 1a) If α ≠ 0 , (1) has no solutions; and nor has the initial system. 1b) If α = 0, then any pair (x, y) that satisfies the second equation in (1), is a solution. Case 2: sin ⎛⎜ β 2 ⎞⎟ ≠ 0 ⇔ β ≠ 2mπ, mεΖ . ⎝

⎠

The system (1) is equivalent to, ⎧ ⎛x −y⎞ ⎫ α ⎪⎪cos⎜ 2 ⎟ = ⎪ ⎝ ⎠ 2 sin ⎛⎜ β ⎞⎟ ⎪ ⎨ ⎝ 2 ⎠ ⎬⎪ ⎪ ⎩⎪x + y = β ⎭⎪

(2)

481

α α ≤ 1 ⇔ α ≤ 2 sin ⎛⎜ β ⎞⎟ ; then there exists a ≤1⇔ ⎝ 2⎠ β β ⎛ ⎞ ⎛ ⎞ 2 sin ⎜ 2 sin ⎜ ⎟ ⎟ 2 2 ⎝ ⎠ ⎝ ⎠ α . We then have, unique angle τ such that 0 ≤ τ ≤ π and cos τ = 2 sin ⎛⎜ β ⎞⎟ ⎝ 2⎠ ⎫ ⎧ ⎛x − y⎞ ⎟ = cos τ⎪ ⎪cos⎜ (2) ⇔ ⎨ ⎝ 2 ⎠ ⎬⇔ ⎪ ⎪x + y = β ⎭ ⎩

2a) If − 1 ≤

⎧x − y ⎫ = 2kπ ± τ⎪ ⎧ x − y = 4kπ ± 2τ⎫ ⎛ the plus sign choice; ⎞ ⎪ ⎟⎟ ⇔⎨ 2 ⎬⇔⎨ ⎬ ⇔ ⎜⎜ + = β x y and the minus choice ⎠ ⎩ ⎝ ⎭ ⎪⎩x + y = β; kεΖ ⎪⎭ ⇔ ⎛⎜ ⎛⎜ x = 2kπ + τ + β ∧ y = −2kπ − τ + β ⎞⎟ ∨ ⎛⎜ x = 2kπ − τ + β ∧ y = −2kπ + τ + β ⎞⎟ ⎞⎟ 2 2⎠ ⎝ 2 2 ⎠⎠ ⎝⎝

2b) If

α > 1 ⇔ α > 2 sin ⎛⎜ β ⎞⎟ ; there are no solutions. ⎝ 2⎠ β ⎛ ⎞ 2 sin ⎜ ⎟ ⎝ 2⎠

We have the following Let α,βε R and S be the solution set of the system F134

⎧sin x + sin y = α ⎫ ⎨ ⎬ ⎩x + y = β ⎭

1) If α = 0 and β = 2ρπ , for some integer ρ; then S = {(x, y ) xε R , yε R , x + y = β}

2) If α ≠ 0 and β = 2ρπ for some integer ρ; S= 0/ ; no solutions. 3) If β ≠ 2mπ, mεΖ ; and α > 2 sin ⎛⎜ β 2 ⎞⎟ , then S= 0/ ; no solutions. ⎝

⎠

482

4) If β ≠ 2mπ, mεΖ ; and α ≤ 2 sin ⎛⎜ β 2 ⎞⎟ ⎝

⎠

then S = S1 U S2 ; where,

{

β S1 = (x, y )(x, y )ε Rx R , x = 2kπ + τ + 2 ,

F134

}

y = −2kπ − τ + β ; kεΖ ; and, 2

{

β S2 = (x, y )(x, y )ε Rx R , x = 2kπ − τ + 2 ,

}

y = −2kπ + τ + β ; kεΖ 2

and where τ stands for the unique angle such that 0 ≤ τ ≤ π and cos τ =

α 2 sin ⎛⎜ β ⎞⎟ ⎝ 2⎠

Note: When τ=0 or π; the two sets S1 and S2 are actually equal, S1 = S2 . While when 0 2 cos⎛⎜ β ⎞⎟ ; there are no solutions. ⎝ 2⎠ 2 cos⎛⎜ β ⎞⎟ ⎝ 2⎠

α . β ⎛ ⎞ 2 cos⎜ ⎟ ⎝ 2⎠

484

We have the following. Let α, βε R and S be the solution set of the system ⎧sin x − sin y = α ⎫ ⎨ ⎬ ⎩x + y = β ⎭

F135

1) If α=0 and β = 2ρπ + π ; for some integer ρ; then, S = {(x , y ) xε R , yε R , x + y = β}

2) If α ≠ 0 and β = 2ρπ + π , for some ρε Z; then, S= 0/ ; no solutions. 3) If β ≠ 2mπ + π, mεZ ; and α > 2 cos⎛⎜ β 2 ⎞⎟ , then S= 0/ ; no solutions. ⎝

⎠

4) If β ≠ 2mπ + π, mεZ ; and α ≤ 2 cos⎛⎜ β 2 ⎞⎟ , then S = S1 U S2 ; where ⎝

{

β S1 = (x, y )(x, y )ε Rx R , x = 2kπ + θ + 2

⎠

}

y = −2kπ − θ + β ; kεZ , 2

{

β S2 = (x, y )(x, y )ε R , x R , x = 2kπ + π − θ + 2 y = −2kπ − π + θ + β

2

};

where θ is the unique angle such that − π 2 ≤ θ ≤ π 2 and sin θ =

α 2 cos⎛⎜ β ⎞⎟ ⎝ 2⎠

Remark: When θ = − π 2 or π 2 ; S1 = S2 : while for − π 2 < θ < π 2 , S1 I S2 = 0/

485

(iii) Solution of System (iii) ⎧sin x + sin (− z ) = α ⎫ ⎧sin x − sin z = α ⎫ ⎧sin x + sin y = α ⎫ ⎪ ⎪ ⎪ ⎪ We have ⎨ ⎬ ⇔ ⎨x + z = β ⎬ . The first ⎬ ⇔ ⎨x + z = β ⎩x − y = β ⎭ ⎪z = − y ⎪ ⎪z = − y ⎪ ⎩ ⎭ ⎩ ⎭

two equations describe a system in x and z, which is of type (ii), governed by F135. We apply F135 to find the solutions in x and z; and then by using –y=z we find the solutions in x and y. We have the following.

Let α, βε R, and S be the solution set of the system ⎧sin x + sin y = α ⎫ ⎨ ⎬ ⎩x − y = β ⎭ F136

1) If α=0 and β = 2ρπ + π , for some ρεZ ; then S = {(x , y ) xε R , yε R , x − y = β}

2) If α ≠ 0 , β = 2ρπ + π , for some ρεZ ; then S = 0/ . 3) If β ≠ 2mπ + π , mεZ , and α > 2 cos⎛⎜ β ⎞⎟ ; then S = 0/ ⎝ 2⎠ 4) If β ≠ 2mπ + π , mεZ , and α ≤ 2 cos⎛⎜ β ⎞⎟ ; then S = S1 U S2 ; where, ⎝ 2⎠

{

β S1 = (x , y )(x, y )ε Rx R , x = 2kπ + θ + 2

}

y = 2kπ + θ − β ; kεZ , 2

{

β S2 = (x, y )(x, y )ε R , x R , x = 2kπ + π − θ + 2 y = 2kπ + π − θ − β

2

};

where θ is the unique angle with − π 2 ≤ θ ≤ π 2 and

486

F136

sin θ =

α . β ⎛ ⎞ 2 cos⎜ ⎟ ⎝ 2⎠

Remark: For θ = − π 2 or π 2 ; S1 = S2 . While for − π 2 < θ < π 2 , S1 I S2 = 0/

(iv) Solution of System (iv) ⎧sin x + sin z = α ⎫ ⎧sin x − sin y = α ⎫ ⎪ ⎪ We have ⎨ ⎬ . The first two equations describe ⎬ ⇔ ⎨x + z = β ⎩x − y = β ⎭ ⎪z = − y ⎪ ⎩ ⎭

a system in x and z which is of type (i), governed by F134. We apply F134 to find the solutions in x and z; and then by using z = − y, to find the solutions in x and y. We have the following. Let Let α, βε R, and S be the solution set of the system ⎧sin x + sin y = α ⎫ ⎨ ⎬ ⎩x − y = β ⎭ F137

1) If α=0 and β = 2ρπ , for some ρεZ ; then S = {(x , y ) xε R , yε R , x − y = β}

2) If α ≠ 0 , β = 2ρπ , for some ρεZ ; then S = 0/ . 3) If β ≠ 2mπ , mεZ , and α > 2 sin β 4) If β ≠ 2mπ , mεZ , and α ≤ 2 sin β

2

2

; then S = 0/ ; then S = S1 U S2 ; where,

487

{

β S1 = (x, y )(x, y )ε Rx R , x = 2kπ + τ + 2

}

y = 2kπ + τ − β ; kεZ , 2

F137

{

β S2 = (x, y )(x, y )ε R , x R , x = 2kπ − τ + 2

}

y = 2kπ − τ − β ; kεZ ; 2

where τ is the unique angle such that 0 ≤ τ ≤ π and cos τ =

α . β ⎛ ⎞ 2 sin ⎜ ⎟ ⎝ 2⎠

Note: When τ = 0 or π ; then S1 = S2 ; while when 0 ≤ τ ≤ π , S1 I S2 = 0/

(v) Solution of System (v) We have

(

)

(

)

⎧cos π − z + cos π − w = α ⎫ ⎧sin z + sin w = α ⎫ 2 2 ⎪ ⎪ ⎪ ⎪⎪ ⎧cos x + cos y = α ⎫ ⎪ ⎪ ⎪ ⎨ ⎬ ⇔ ⎨x + y = β ⎬ ⇔ ⎨z + w = π − β ⎬. ⎩x + y = β ⎭ ⎪ ⎪ ⎪ ⎪ π π ⎪⎩z = π 2 − x , w = π 2 − y⎪⎭ ⎪⎩z = 2 − x , w = 2 − y ⎪⎭

The first two equations describe a type (i) system in z and w and with π-β instead of β. By solving this system in accordance with F134; we obtain the solutions in z and w; and from x = π 2 − z, y = π 2 − w , we obtain the solutions to the initial system. We have the following.

488 Let Let α, βε R, and S be the solution set of the system ⎧cos x + cos y = α ⎫ ⎨ ⎬ ⎩x + y = β ⎭ F138

1) If α=0 and β = −2ρπ + π , for some integer ρ; then S = {(x, y ) xε R , yε R , x + y = β}

2) If α ≠ 0 , β = −2ρπ + π , for some ρεZ ; then S = 0/ . ⎛π−β⎞ β 3) If β ≠ −2mπ + π , mεZ , and 2 sin ⎜ ⎟ = 2 cos⎛⎜ 2 ⎞⎟ < α ; ⎝ ⎠ ⎝ 2 ⎠ then S = 0/ 4) If β ≠ −2mπ + π , mεZ , and α ≤ 2 cos β

{

2

; then S = S1 U S2 ; where,

β S1 = (x, y )(x, y )ε Rx R , x = −2kπ + π 2 − τ − 2

}

y = 2kπ + π + τ − β ; kεZ , 2 2

{

β S2 = (x, y )(x, y )ε R , x R , x = −2kπ + π 2 + τ − 2

}

y = 2kπ + π − τ − β ; kεZ ; 2 2

where τ is the unique angle such that 0 ≤ τ ≤ π and cos τ =

α α . = ⎛ π − β ⎞ 2 cos⎛⎜ β ⎞⎟ 2 sin ⎜ ⎟ ⎝ 2⎠ ⎝ 2 ⎠

Remark: If τ = 0 or π ; then S1 = S2 ; And if 0 2 sin ⎛⎜ 2 ⎞⎟ , ⎝ ⎠ ⎝ 2 ⎠ then S = 0/ 4) If β ≠ −2mπ , mεZ , and α ≤ 2 sin ⎛⎜ β ⎞⎟ ; then S = S1 U S2 ; where, ⎝ 2⎠

{

β S1 = (x, y )(x, y )ε Rx R , x = −2kπ + π 2 − θ − 2

}

y = 2kπ + π + θ − β ; kεZ , 2 2

490

{

β S2 = (x, y )(x, y )ε R , x R , x = −2kπ − π 2 + θ − 2

}

y = 2kπ + 3π − θ − β ; kεZ ; 2 2

where θ is the unique angle such that − π 2 ≤ θ ≤ π 2 and sin θ =

α α = . ⎛ π − β ⎞ 2 sin ⎛⎜ β ⎞⎟ 2 cos⎜ ⎟ ⎝ 2⎠ ⎝ 2 ⎠

Note: For θ = − π 2 or π 2 ; S1 = S2 . Otherwise, S1 I S2 = 0/ .

(vii) Solution of System (vii) ⎧cos x + cos z = α ⎫ ⎧cos x + cos y = α ⎫ ⎪ ⎪ We have ⎨ ⎬ ⇔ ⎨x + z = β ⎬ . The first two equations of the ⎭ ⎩x − y = β ⎪y = −z ⎪ ⎩ ⎭

last system describe a type (v) system in x and z, which is solved in accordance with F138. After finding the solutions in x and z, by using y = −z ; we determine the solutions to the original system. We have the following. Let α, βε R, and S be the solution set of the system ⎧cos x + cos y = α ⎫ ⎨ ⎬ ⎩x − y = β ⎭ F140

1) If α=0 and β = −2ρπ + π , for some ρεZ ; then S = {(x, y ) xε R , yε R , x + y = β}

2) If α ≠ 0 , and β = −2ρπ + π , for some ρεZ ; then S = 0/ .

491 ⎛ π−β⎞ 3) If β ≠ −2mπ + π , mεZ , and 2 cos⎜ ⎟ < α , then S = 0/ ⎝ 2 ⎠ 4) If, β ≠ −2mπ + π , mεZ , and; α ≤ 2 cos⎛⎜ β ⎞⎟ then S = S1 U S2 ; where, ⎝ 2⎠

F140

{

β S1 = (x, y )(x, y )ε Rx R , x = −2kπ + π 2 − τ − 2

}

y = −2kπ − π − τ + β ; kεZ , 2 2

{

β S2 = (x, y )(x, y )ε R , x R , x = −2kπ + π 2 + τ − 2

}

y = −2kπ − π − τ + β ; kεZ ; 2 2

where τ is the unique angle such that 0 ≤ τ ≤ π and cos τ =

α . 2 cos⎛⎜ β ⎞⎟ ⎝ 2⎠

Remark: For τ = 0 or π ; S1 = S2 . Otherwise, S1 I S2 = 0/ .

(viii) Solution of System (viii) ⎧cos x − cos z = α ⎫ ⎧cos x − cos y = α ⎫ ⎪ ⎪ We have ⎨ ⎬ ⇔ ⎨x + z = β ⎬ . The first two equations of the ⎭ ⎩x − y = β ⎪ y = −z ⎪ ⎩ ⎭

last system describe a type (vi) system in x and z, which is solved in accordance with F139. After finding the solutions in x and z, by using y = −z ; we determine the solutions to the original system. We have the following.

492 Let α, βε R, and S be the solution set of the system ⎧cos x − cos y = α ⎫ ⎨ ⎬ ⎩x − y = β ⎭ F141

1) If α=0 and β = −2ρπ , for some ρεZ ; then S = {(x, y ) xε R , yε R , x + y = β}

2) If α ≠ 0 , and β = −2ρπ , for some ρεZ ; then S = 0/ . 3) If β ≠ −2mπ , mεZ , and α > 2 sin ⎛⎜ β ⎞⎟ , then S = 0/ ⎝ 2⎠ 4) If, β ≠ −2mπ , mεZ , and; α ≤ 2 sin ⎛⎜ β ⎞⎟ then S = S1 U S2 ; where, ⎝ 2⎠

{

β S1 = (x, y )(x, y )ε Rx R , x = −2kπ + π 2 − θ − 2

}

y = −2kπ − π − θ + β ; kεZ , 2 2

{

β S2 = (x, y )(x, y )ε R , x R , x = −2kπ − π 2 + θ − 2

}

y = −2kπ − 3π + θ + β ; kεZ ; 2 2

where θ is the unique angle with − π 2 ≤ θ ≤ π 2 and sin θ =

α . 2 sin ⎛⎜ β ⎞⎟ ⎝ 2⎠

Note: For θ = − π 2 or π 2 ; S1 = S2 . Otherwise, S1 I S2 = 0/ .

493

Group 2: Analysis and determination of Solution Sets As we shall see, it is only necessary to solve the first system, the other five reduce to the first.

(i) Solution of System (i) ⎧sin x sin y = α ⎫ ⎧2 sin x sin y = 2α ⎫ ⎬⇔⎨ ⎬⇔ ⎩x + y = β ⎭ ⎩x + y = β ⎭

We have, ⎨

⎧cos(x − y ) − cos(x + y ) = 2α ⎫ ⎧cos(x − y ) = 2α + cos β⎫ ⇔⎨ ⎬⇔⎨ ⎬ ⎩x + y = β ⎭ ⎩x + y = β ⎭

(1)

If 2α + cos β > 1 , the first equation in (1) has no solutions, in which case there are no solutions to the original system. On the other hand, if 2α + cos β ≤ 1 ⇔ 1 − cos β ⎛ 1 + cos β ⎞ ⇔ −1 ≤ 2α + cos β ≤ 1 ⇔ −⎜ ; then, let ω be the unique angle ⎟≤α≤ 2 2 ⎝ ⎠

such that 0 ≤ ω ≤ π and cos ω = 2α + cos β ; and so, ⎧cos(x − y ) = cos ω⎫ ⎧x − y = 2kπ ± ω⎫ (1) ⇔ ⎨ ⎬⇔⎨ ⎬⇔ ⎩x + y = β ⎭ ⎩x + y = β ⎭ ⎛ ⎧x − y = 2kπ + ω⎫ ⎧x − y = 2kπ − ω⎫ ⎞ (1) ⎜⎜ ⎨ ⎬ ⎟⎟ ⇔ ⎬∨ ⎨ x y x y + = β + = β ⎭⎠ ⎭ ⎩ ⎝⎩ ⎞ ⎛⎧ β−ω ⎫ ω + β ⎞⎫ ⎧ ⎟ ⎜ ⎪x = kπ + ⎛⎜ ⎟ ⎪ ⎪ x = kπ + ⎪ 2 ⎪ ⎟ ⎜⎪ ⎝ 2 ⎠⎪ ∨ ⎪ ⎬ ⎨ ⎬ ; kεΖ ⎟ ⎜⎨ β − ω ⎪ ⎪ y = − kπ + ⎛ β + ω ⎞ ⎪ ⎜ ⎟ ⎟ ⎜ ⎪ y = − kπ + ⎝ 2 ⎠⎪⎭ 2 ⎪⎭ ⎪⎩ ⎠ ⎝ ⎪⎩

494

We have the following. Let α, βε R, and S be the solution set of the system ⎧sin x sin y = α ⎫ ⎨ ⎬ ⎩x + y = β ⎭

F142

1) If 2α + cos β > 1 , the system has no solutions; S = 0/ 1 − cos β ⎛ 1 + cos β ⎞ , then ⎟≤α≤ 2 2 ⎝ ⎠

2) If 2α + cos β ≤ 1 or equivalently, − ⎜ S = S1 U S2 ; where

⎧ ⎛ω+β⎞ ⎟ S1 = ⎨(x , y )(x, y )ε Rx R , x = kπ + ⎜ ⎝ 2 ⎠ ⎩ β−ω ⎫ y = −kπ + ; kεZ ⎬ , 2 ⎭ ⎧ ⎛β − ω⎞ and S2 = ⎨(x , y )ε R , x R , x = kπ + ⎜ ⎟ ⎝ 2 ⎠ ⎩ ⎫ ⎛β + ω⎞ y = − kπ + ⎜ ⎟; kεZ ⎬ ;0 ≤ ω ≤ π, cos ω = 2α + cos β ⎝ 2 ⎠ ⎭

Note: When ω = 0 or π ; S1 = S2 . Otherwise, S1 I S2 = 0/ .

(ii) Solution of System (ii) ⎧sin x sin (− z ) = α ⎫ ⎧sin x sin y = α ⎫ ⎪ ⎪ We have, ⎨ ⎬ ⇔ ⎨x + z = β ⎬⇔ ⎩x − y = β ⎭ ⎪y = −z ⎪ ⎩ ⎭

495 ⎧sin x sin z = −α ⎫ ⎪ ⎪ ⇔ ⎨x + z = β ⎬ The first two equations of the last system describe a type (i) ⎪ y = −z ⎪ ⎩ ⎭

system in x and z, but with (-α) in place of α. This is solved according to F142 and subsequently we use y=-z to find the solutions of the initial system. We have the following. Let α, βε R, and S be the solution set of the system ⎧sin x sin y = α ⎫ ⎨ ⎬ ⎩x − y = β ⎭

F143

1) If − 2α + cos β > 1 , S = 0/ ⎛ cos β − 1 ⎞ ⎛ 1 + cos β ⎞ ⎟≤α≤⎜ ⎟ ; then, S = S1 U S2 , 2 2 ⎝ ⎠ ⎝ ⎠

2) If − 2α + cos β ≤ 1 ⇔ ⎜ with

⎧ ⎛ω+β⎞ ⎟ S1 = ⎨(x , y )(x, y )ε Rx R , x = kπ + ⎜ ⎝ 2 ⎠ ⎩ ω−β ⎫ y = kπ + ; kεZ ⎬ , 2 ⎭ ⎫ ⎧ ⎛β − ω⎞ ⎛β + ω⎞ ⎟, y = kπ − ⎜ ⎟; kεZ ⎬ S2 = ⎨(x , y )(x, y )ε R , x R , x = kπ + ⎜ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎭ ⎩

where ω is the unique angle such that 0 ≤ ω ≤ π and cos ω = −2α + cos β

Remark: When ω = 0 or π ; S1 = S2 ; otherwise, S1 I S2 = 0/ .

496

(iii) Solution of System (iii)

(

)

⎧sin x cos π − z = α ⎫ 2 ⎪ ⎪ ⎧sin x cos y = α ⎫ ⎪ ⎪ π We have, ⎨ ⎬ ⇔ ⎨x + 2 − z = β ⎬⇔ ⎩x + y = β ⎭ ⎪ ⎪ ⎪⎩z = π 2 − y ⎪⎭

(

)

⎧sin x sin z = α ⎫ ⎪ ⎪ ⎪ ⎪ π ⇔ ⎨x − z = β − ⎬ The first two equations of the last system describe a type (ii) 2 ⎪ ⎪ ⎪⎭ ⎪⎩ y = π 2 − z

(

)

system in x and z, but with β − π 2 in place of β. This is solved according to F143 and subsequently we use y = π 2 − z to find the solutions to the initial system. Note

(

)

(

)

that cos β − π 2 = cos π 2 − β = sin β . We have the following. Let α, βε R, and S be the solution set of the system ⎧sin x cos y = α ⎫ ⎨ ⎬ ⎩x + y = β ⎭

F144

(

)

1) If − 2α + cos β − π 2 > 1 ⇔ − 2α + sin β > 1 ; S = 0/ 2) If − 2α + sin β ≤ 1 ⇔

sin β − 1 1 + sin β ≤α≤ ; S = S1 U S2 , where 2 2

⎧ ⎛β + ω⎞ ⎟ S1 = ⎨(x , y )(x, y )ε Rx R , x = kπ − π 4 + ⎜ ⎝ 2 ⎠ ⎩ ⎫ ⎛ω−β⎞ ; kε Z ⎬ , y = kπ + π + ⎜ 4 ⎝ 2 ⎟⎠ ⎭

497

F144

⎧ ⎛β − ω⎞ ⎟, S2 = ⎨(x , y )(x , y )ε R , x R , x = kπ − π 4 + ⎜ ⎝ 2 ⎠ ⎩ ⎫ ⎛β + ω⎞ ; kεZ ⎬ y = kπ + π − ⎜ ⎟ 4 ⎝ 2 ⎠ ⎭

where ω is the unique angle such that 0 ≤ ω ≤ π and cos ω = −2α + sin β .

Note: When ω = 0 or π ; S1 = S2 . Otherwise, S1 I S2 = 0/ .

(iv) Solution of System (iv) ⎧sin x cos y = α ⎫ ⎬ By introducing the substitution z=-y, the ⎩x − y = β ⎭

We have the system ⎨

given system reduces to a type (iii) system in x and z. We just state the result. Let α, βε R, and S be the solution set of the system ⎧sin x cos y = α ⎫ ⎨ ⎬ ⎩x − y = β ⎭

F145

1) If − 2α + sin β > 1 , S = 0/ 1 + sin β ⎛ sin β − 1 ⎞ ; S = S1 U S2 , with ⎟≤α≤ 2 ⎝ 2 ⎠

2) If − 2α + sin β ≤ 1 ⇔ ⎜

⎧ ⎛β + ω⎞ ⎟ S1 = ⎨(x , y )(x, y )ε Rx R , x = kπ − π 4 + ⎜ ⎝ 2 ⎠ ⎩ ⎫ ⎛β − ω⎞ ; kε Z ⎬ , y = − kπ − π + ⎜ ⎟ 4 ⎝ 2 ⎠ ⎭

498 ⎧ ⎛β − ω⎞ ⎟, S2 = ⎨(x , y )(x , y )ε R , x R , x = kπ − π 4 + ⎜ ⎝ 2 ⎠ ⎩ ⎫ ⎛β + ω⎞ ; kεZ ⎬ y = − kπ − π + ⎜ ⎟ 4 ⎝ 2 ⎠ ⎭

F145

where ω is the unique angle with 0 ≤ ω ≤ π and cos ω = −2α + sin β .

Remark: For ω = 0 or π ; S1 = S2 ; while for 0 1 , S = 0/ 2) If 2α − cos β ≤ 1 ⇔ S = S1 U S2 ; where,

cos β − 1 1 + cos β ≤α≤ ; 2 2

500 ⎧ ⎛β + ω⎞ ⎟ S1 = ⎨(x , y )(x , y )ε Rx R , x = − kπ + π 2 − ⎜ ⎝ 2 ⎠ ⎩ ⎫ ⎛ω−β⎞ ; kε Z ⎬ , y = − kπ − π + ⎜ ⎟ 2 ⎝ 2 ⎠ ⎭ ⎧ ⎛ω−β⎞ ⎟, S2 = ⎨(x , y )(x , y )ε R , x R , x = − kπ + π 2 + ⎜ ⎝ 2 ⎠ ⎩

F147

⎫ ⎛β + ω⎞ ; kεZ ⎬ y = − kπ − π + ⎜ ⎟ 2 ⎝ 2 ⎠ ⎭

with 0 ≤ ω ≤ π and cos ω = 2α − cos β . Note: For ω = 0, π ; S1 = S2 . And for 0 < ω < π , S1 I S2 = 0/ .

Group 3: Analysis and determination of Solution Sets Of the eight systems listed, six reduce to (i) or (v). We will only determine the solution sets of systems (i) and (v). The other six systems are listed among the unsolved problems. But first let us see how those six systems reduce to systems (i) or (v). System (ii) reduces to (i) by introducing the substitution z = − y . System (iii) also reduces to system (i) by introducing the substitutions z = π − x, w = π − y , and with π − β in the place of β. System (iv) reduces to 2 2

(iii) (which in turn reduces to (i)) by setting z=-y. System (vi) reduces to (v) by putting z=-y. System (vii) reduces to (vi) (which in turn reduces to (v)) by setting z = π − y and with β − π in place of β. And system (viii) reduces to (vii) by 2 2

putting y=-z.

501

(i) Solution of System (i) ⎧ sin x ⎫ = α⎪ ⎪ The given system is ⎨ sin y ⎬ . As it will become evident below, there is a ⎪x + y = β ⎪ ⎩ ⎭

certain complexity to this system. First observe that any pair (x,y) which is a solution to this system, must satisfy sin y ≠ 0 ⇔ y ≠ mπ, mεΖ . We have, ⎧ sin x ⎫ ⎧ sin x = α⎪ = α, y ≠ mπ, mεΖ ⎪ ⎪ ⎨ sin y ⎬ ⇔ ⎨ sin y ⎪x + y = β ⎪ ⎪x + y = β ⎩ ⎭ ⎩

⎫ ⎪ ⎬ (1) ⎪ ⎭

Case 1: β = ρπ for some fixed integer ρ. ⎧ sin x = α, y ≠ mπ, mεΖ ⎪ We have (1) ⇔ ⎨ sin y ⎪ y = ρπ − x ⎩

⎫ ⎪ ⎬ ⎪ ⎭

(2)

Note that y = ρπ − x ⇔ x = ρπ − y ⇔ sin x = (−1)ρ+1 • sin y . Thus, ⎧ (−1)ρ+1 • sin y = α, y ≠ mπ, mεΖ (2) ⇔ ⎪⎨ sin y ⎪ y = ρπ − x ⎩ ⎧(−1)ρ+1 = α, y ≠ mπ, mεΖ ⎫ ⇔⎨ ⎬ ⎩ y = ρπ − x ⎭

⎫ ⎪ ⎬⇔ ⎪ ⎭ (3)

Clearly the last system shows that if α ≠ −1, 1 , there are no solutions. If ρ is an even integer and α=1,there are no solutions either; likewise if ρ is odd and α= − 1 there are no solutions. If on the other hand ρ is even and α=-1, the solutions to the

502

given equation are all the pairs (x,y) which satisfy x + y = ρπ = β and y ≠ mπ, mεΖ ; while if ρ is an odd integer and α=1, the solutions are again all pairs (x,y) which satisfy x+y=ρπ=β and with y≠mπ, mεZ . Case 2: β = ρπ + π 2 , for some fixed integer ρ. ⎧ sin x = α, y ≠ mπ, mεΖ ⎪ We have (1) ⇔ ⎨ sin y ⎪x = ρπ + π − y 2 ⎩

(

⎫ ⎪ ⎬ ⎪ ⎭

(4)

)

Note that sin x = sin ρπ + π 2 − y = cos(ρπ − y ) = (−1)ρ • cos(− y ) ; ⎧ (−1)ρ • cos y = α, y ≠ mπ, mεΖ ⎪⎪ sin x = (−1)ρ cos y . Therefore, (4) ⇔ ⎨ sin y ⎪x = ρπ + π − y ⎪⎩ 2

⎫ ⎪⎪ ⎬ ; and since ⎪ ⎪⎭

(−1)ρ cos y = α ⇔ (−1)2ρ cos y = (−1)ρ • α; cos y = (−1)ρ • α ; the last system is equivalent ⎧⎪cot y = (−1)ρ • α, y ≠ mπ, mεΖ to, ⎨ ⎪⎩x = ρπ + π 2 − y

⎫⎪ ⎬ ⎪⎭

(5)

Let φ be the unique angle such that 04α. The second condition gives 4f(1)>0 ⇔ f(1)>0 ⇔ 4-4α-β2>0 ⇔ 2α4α^4α