# If sin(2x +3y)=1; cos(2x - 3y)= √3/ 2, find x and y

2
by Rajjum

2015-09-15T13:15:46+05:30
Given:
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
2015-09-15T15:24:49+05:30

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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
=> general solution:  2x - 3y = 2 m π + π/6   or  2 m π - π/6  ---- (3)    m = integer
principal solution:  2x - 3y = π/6  or  -π/6  or 11π/ 6    --- (4)

Solving  (2) and (4) together:  for principal solutions:
4x = 4π/6  or  2π/6  or  14π/6
=>  x = π/6  or  π/12  or  7π/12    --- (5)
So,  y = [π/2 - 2x]/3 = π/6 - 2x/3 =  π/18  or  π/9  or  -2π/9  ---(6)

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

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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

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