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## Answers

sin (2x+3y)=1 and cos (2x-3y)=√3 /2

thus 2x+3y=90 (bcoz sin90 =1)

and 2x-3y=30

solving both the eqns,we get:

2x+3y=90

2x -3y=30

--------------

4x+0 =120

thus x=30.

substitute x in any eqn, we get y=10

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Sin (2x + 3y) = 1

=> 2x + 3y = 2n π + π/2 --- (1) n = integer

principal solution 2x + 3 y = π/2 --- (2)

Cos (2x - 3y) = √3 /2

=>

Solving (2) and (4) together: for principal solutions:

4x = 4π/6 or 2π/6 or 14π/6

=>

So,

===================================

we will get more general solutions using (1) and (3)...

m and n are integers.

4x = 2(n+m)π+ 2π/3

y = [2nπ +π/2 - 2x ]/3 =>

Also, 4x = 2(n+m)π + 7π/3 =>

then

similarly we get also:

=> 2x + 3y = 2n π + π/2 --- (1) n = integer

principal solution 2x + 3 y = π/2 --- (2)

Cos (2x - 3y) = √3 /2

=>

**2x - 3y = 2 m π + π/6 or 2 m π - π/6 ---- (3) m = integer***general solution:***2x - 3y = π/6 or -π/6 or 11π/ 6 --- (4)***principal solution:*Solving (2) and (4) together: for principal solutions:

4x = 4π/6 or 2π/6 or 14π/6

=>

**--- (5)***x = π/6 or π/12 or 7π/12*So,

**[π/2 - 2x]/3 = π/6 - 2x/3 =***y =***---(6)***π/18 or π/9 or -2π/9**(x,y): = (π/6, π/18) or (π/12, π/9) or (7π/12, -2π/9)*===================================

we will get more general solutions using (1) and (3)...

m and n are integers.

4x = 2(n+m)π+ 2π/3

*x = (n+m)π/2 + π/6*y = [2nπ +π/2 - 2x ]/3 =>

*y = (n-m)π/3 -π/18*Also, 4x = 2(n+m)π + 7π/3 =>

*x = (n+m) π/2 + 7π/12*then

*y = (n-m)π/3 - 2π/9]*similarly we get also:

*x = (n+m)π/2 + π/12 and y = (n-m)π/3 + π/9*