Question

sinx If = 3 and sin y value of 116 COS X cosy 2' sin 2x cos2x sin 2y cos2y]​

Answer 1

Answer:

x=nπ+(−1)

n

6

π

andy=2nπ±

3

2π

Step-by-step explanation:

2

sinx+cosy

=1=2

0

⇒sinx+cosy=0…(1)

and 16

sin

2

x+cos

2

y

=4=16

2

1

sin

2

x+cos

2

y=

2

1

…(2)

Subsitute cosy=−sinx in equation (2) from equation (1)

sin

2

x+sin

2

x=

2

1

2sin

2

x=

2

1

⇒sinx=±

2

1

Now sinx=

2

1

⇒cosy=−

2

1

⇒x=nπ+(−1)

n

6

π

andy=2nπ±

3

2π

and sinx=−

2

1

⇒cosy=

2

1

x=nπ+(−1)

n+1

6

π

andy=2nπ±

3

π