Find all values of x such that sin 2x = sin x and 0 <= x <= 2π. (List the answers in increasing order.)

**Answer 1**

sin2x = sinx OR sinx(2cosx — 1) = 0 giving sinx = sin0, cosx = cos(pi/3)

whence x = 0, pi/3, pi, 5pi/3, 2pi.

**Answer 2**

sin 2x = sin x

=> sin 2x – sin x = 0

=> 2 sin x cos x = sin x = 0

=> sin x (2 cos x – 1 ) = 0

=> sin x = 0 or 2cos x – 1 = 0

sin x = 0

=> x = 0, pi

2 cos x – 1 = 0

=> cos x = 1/2

=> x = pi/3, 2pi/3, 4pi/3, 5pi/3

have a nice day!

**Answer 3**

trig identity:

sin 2x =2sinxcosx

so you want x s.t.

2sinxcosx=sinx

this is true when sinx=0 or when cosx=1/2.

sin(0)=sin(2pi)=0,

cos(pi/3)=cos(5pi/3)=1/2

so you want 0 ,pi/3 ,pi, 5pi/3 or 2pi

**Answer 4**

0

pi/3

2pi/3

pi

2pi

You might want to check it with a calculator, we don’t have one. jk – we’re too lazy to use it.