Question

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

Answer 1

Solution:

i)sin(2x+3y) = 1

=> sin(2x+3y) = sin90°

=> 2x+3y = 90 ----(1)

ii) cos(2x-3y) = √3/2

=> cos(2x-3y) = cos30°

=> 2x-3y = 30 ---(2)

Add (1)&(2) , we get

=> 4x = 120

=> x = 120/4

=> x = 30

iii ) Put x = 30 in equation (1),

we get

2×30+3y = 90

=> 60+3y = 90

=> 3y = 90-60

=> 3y = 30

=> y = 30/3

=> y = 10

Therefore,

x = 30, y = 10

••••

Answer 2

$\textbf{\underline{\underline{According\:to\:the\:Question}}}$

sin (2x + 3y) = 1

⇒ sin (2x + 3y) = sin 90°

⇒ 2x + 3y = 90 [Consider equation (1)]

Now we have :-

${\implies cos (2x - 3y) = \dfrac{\sqrt{3}}{2}}$  

⇒ cos (2x - 3y) = cos 30°

⇒ 2x - 3y = 30 [Consider equation (2)]

Adding Equation 1 and 2 we get values :-

⇒ 4x = 120

$\tt{\rightarrow x = \dfrac{120}{4}}$    

⇒ x = 30

Hence we got the value of x put the value of x in equation (1) we get values

⇒ 2 × 30 + 3y = 90

⇒ 60 + 3y = 90

⇒ 3y = 90 - 60

⇒ 3y = 30

$\tt{\rightarrow y = \dfrac{30}{3}}$    

⇒ y = 10

Hence we get the values

${\boxed{\bigstar{{x = 30\:and\:y=10}}}}$