Question

Find all solutions in the interval [0, 2π).

(sin x)(cos x) = 0

A. pi/2, pi
B. 0, pi/2, pi, 3pi/2
C. pi, 3pi/2
D. 0, 3pi/2

Answer 1

hint
sin x cos x = sin(2x)/2

Answer 2

So either \(\sin x = 0\) or \(\cos x=0\).

So you know where in unit circle ( interval [0, 2π) ) is where \(\sin x = 0\) or \(\cos x=0\)?

Answer 3

(sin x)(cos x) = 0

has the same solutions as
sin(2x)/2 = 0

sin(2x) = 0

x = arcsin(0)/2

Answer 4

So, sin x = 0?

Answer 5

nope

Answer 6

sin(2x) = 0 , theres a difference

Answer 7

Sinx =0 at 0 and also pi and multples of pi. Cosx =0 at pi/2 and odd multiples of pi/2. So the answer is B?