Find all values of x such that sin 2x = sin x and 0 <= x <= 2π. ?

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.

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