Question

If sin(3x+2y)=1 and cos(3x-2y)=√3/2 find the value of x and y

Answer 1

sin(3x+2y)=1              (using trignometric table we get sin90=1)
3x + 2y = 90      -(1)
cos (3x-2y)=√3/2         (using trignometric table we get cos30=√3/2)
3x - 2y = 30        -(2)
Adding equation 1 and equation 2, we get
(3x + 2y) + (3x - 2y) = 90 +30
3x + 3x + 2y - 2y =120
6x = 120
x= 20
Now putting the value of x in equation 1, we get
3(20) + 2y =90
60 + 2y =90
2y = 90 - 60
2y = 30
y = 15