Question

sin x cos x=0
$0 \leqslant x \leqslant 2\pi$
​

Answer 1

Answer:

0,π/2,π,3π/2,2π

Step-by-step explanation:

sinx cosx=0

2sinx cosx=0

sin2x =0

now

0<x<2π

- -

0<2x<4π

- -

now

sin2x=0

2x=0,π,2π,3π,4π

π 2π 3π 4π

x=0,-------,-------,-------,------

2 2 2 2

π 3π

x=0,-------,π--------,2π

2 2

Answer 2

Answer:

Step-by-step explanation:

sinx cosx=0

2sinx cosx=0

sin2x =0

now

0<x<2π

- -

0<2x<4π

- -

now

sin2x=0

2x=0,π,2π,3π,4π

π 2π 3π 4π

x=0,-------,-------,-------,------

2 2 2 2

π 3π

x=0,-------,π--------,2π

2 2