what is the radians of sin x = sin 2x?

Question

dealing with calculus

0 to 2pi is the actual radian scale

Answer 1

For the right side, use the trigonometric identity sin 2x = 2 sin x cos x. This transforms the equation into

sin x = 2 sin x cos x

Subtract 2 sin x cos x from both sides:

sin x - 2 sin x cos x = 0

Factor out sin x from the left:

sin x(1 - 2 cos x) = 0

So the solutions are where sin x = 0 or where 1 - 2 cos x = 0; that is, where sin x = 0 or cos x = 1/2.

On the interval [0, 2π), the x-values where sin x = 0 are 0 and π, and the x-values where cos x = 1/2 are π/3 and 5π/3.

So 0, π/3, π, and 5π/3 are solutions.

But the sine and cosine functions are periodic with period 2π, so we will get more solutions if we add any multiple of 2π.

So the solutions are 2kπ, π/3 + 2kπ, π + 2kπ, and 5π/3 + 2kπ, where k is any integer.

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EDIT: If you are only interested in solutions in the interval [0, 2π], including 2π, the solutions are

0, π/3, π, 5π/3, 2π.

If you are only interested in solutions in the interval [0, 2π), not including 2π, then the solutions are

0, π/3, π, 5π/3.

Answer 2

sin 2x - sin x = 0
2 sin x cos x - sin x = 0
sin x (2 cos x - 1) = 0
sin x = 0 , cos x = 1/2
x = 0 , π , 2π , π/3 , 5π/3

Answer 3

Sins = Sin 2x
or, Sinx -2SinxCosx=0
or, Sinx(1-2Cosx)=0
or, SinX =0
so, X=nPi
Also, (1-2CosX)=0
or, CosX=1/2=Cos60
X=60 deg
180 deg = pi radians
60 deg (pi/180)*60 = pi/3 radians
So pi/3 +2n*pi radians and n*pi rad (n=0,1,2...)