# Solve the equation sinx=sin2x x is in [0;2pi)

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### 3 Answers

sinx = sin2x

We know that sin2x = 2sinxcosx

==> sin x = 2sinx cosx

==> sinx - 2sinx*cosx = 0

Now factor sinx:

==> sinx (1- 2cosx) = 0

==> sinx = 0 ==> x1 = 0, pi

==> 1-2cosx = 0==> cosx = 1/2 ==> x2= pi/3, 5pi/3

==> x = { 0, pi, pi/3, 5pi/3}

To solve the equation, we'll apply the formula:

sin x = 2 sin x *cos x (1)

We'll substitute (1) in the given equation:

sin x - sin 2x=0

sin x - 2 sin x *cos x=0

We'll factorize by sin x:

sin x*(1- 2cos x)=0

We'll set the factors as 0:

sin x=0 or cos x =1/2

sin x =0 and x belongs to [0,2pi)

x=(-1)^k*arcsin0 + k*pi

**x={0,pi}**

cos x =1/2

x=+/-arccos (1/2)+2*k*pi

**x={pi/3, 5pi/3}**

To solve sinx =sin2x in [0,2pi]

Solution

Sin2x = 2sinxcosx is a triginometric identity. Substitute this in the goven equation:

sinx = 2sinx*cosx

Sinx-2sinxcosx = 0

sinx(1-2cosx) = 0

Sinx = 0 Or 1-2cosx =0

sinx = 0 gives x = 0 or pi in the x belonging to (0,2pi)

1-2cosx = 0 gives. cosx =1/2. Or x = pi/3 or x = - pi/3 Or 5pi/3

So x = 0, pi/3,pi ,5pi/3