Question

Q3: If 4cos^3y-1=0, find the value of y.​

Answer 1

Answer:

3x−2y=5............ (1)

3y−2x=3

−2x+3y=3

Multiply (1) by 2 & eq (2) by 3

6x−4y=10..........(3)

−6x+9y=19..........(4)

_______________

Adding eqn (3) and (4), we get

6x−4y=10

−6x+9y=19

___________

5y=1

y=

5

1

Putting the value in

6x−

5

4

=10

6x=10+

5

4

=

5

54

x=

5×6

54

=

5

9

x=

5

9

Now x+y=

5

9

+

5

1

=

5

10

=2$$

x+y=2