If sin(2x+3y)=1and cos(2x-3y)=root 3÷2..find the value of x and y

1.	If sin(2x+3y)=1and cos(2x-3y)=root 3÷2..find the value of x and y
Answer» 2x+3y=90. (Sin 90°=1). (1)2x-3y=√3/2 (cos 30°=√3/2). (2) Solving both the equation, we get 2x+3y=902x-3y=30= 4x=120= x=120/4= x= 30Now, Putting the value of x=30 in (1)2×30+3y=9060+3y=903y=30Y=10Therefore, the value of x=30 and y=10.

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