# Solve for x : sin3x=2sin^3x

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### 4 Answers

sin3x = 2sin^3 x

First we will rewrite:

sin3x = sin(2x+x)

But we know that :

sin(A+B) = sinAcosB + cosAsinB

==> sin(2x+x) = sin2xcosx + sinxcos2x

But sin2x = 2sinxcosx

==> sin3x = 2sinxcosxcosx + sinx (1-2sin^2 x)

= 2sinxcos^2x + sinx ( 1- 2sin^2 x)

=2sinx ( 1-sin^2 x) + sinx ( 1- 2sin^2 x)

= 2sinx -2sin^3 x + sinx - 2sin^3 x

==> sin3x = 3sinx - 4sin^3 x = 2sin^3 x

==> 3sinx = 6sin^3 x

Divide by 3

==> sinx = 2sin^3 x

==> 2sin^3 x - sinx = 0

==> sinx( 2sin^2 x -1) = 0

==> sinx = 0 ==> x = o, pi, 2pi

==> sin^2 x -1 = 0

==> sin^2 x = 1/2

==> sinx = 1/sqrt2

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

**==> x = { 0, pi/4, 3pi/4, 2pi } + 2npi n= 0, 1, 2, ....**

We have to solve for x given sin 3x = 2*(sin x)^3

Now sin 3x = 3 sin x - 4 (sin x)^3

So we get

3 sin x - 4 (sin x)^3 = 2*(sin x)^3

=> 3 sin x = 6* (sin x)^3

=> 3 = 6* (sin x)^2

=> (sin x)^2 = 3/6

=> (sin x)^2 = 1/2

=> sin x = -1/ sqrt 2 and + 1/sqrt 2

x= arc sin (-1/ sqrt 2 ) and x = arc sin (1/ sqrt 2)

=> x = -45 degrees and +45 degrees.

**Therefore x is equal to 45 degrees and -45 degrees.**

sin3x=2sin^3x.

We know that sin3x = sin(2x+x).

= sin2x*cosx +cos2xsinx

= (2sinxcosx)cosx+(1-2sin^2x)sinx.

= 2sinxcos^2x+sinx-2sin^3x.

= 2sinx(1-sin^2x)+sinx-2sin^3x

= 3sinx-4sin^3x.

Therefore the given expression sin3x=2sin^3x could be written as:

3sinx - 4sin^3 = 2sin^3x.

3sinx -6sin^3x = 0

3sinx(1-2sin^2x) = 0.

sinx(1/2-sin^2x) = 0.

sinx((sinx-1/sqrt2)(sinx-1/sqrt2) = 0.

So sinx = 0, sinx = 1/sqrt2 , or sinx = -1/sqrt2.

x= npi, or x = npi +or- (-1)^n*pi/4.

We know that sin 3x = 3sinx - 4(sin x)^3.

We'll substitute it in the equation:

3sinx - 4(sin x)^3 = 2(sin x)^3

We'll add 4(sin x)^3 both sides:

3sin x = 6(sin x)^3

We'll divide by 3:

sin x = 2(sin x)^3

We'll subtract sin x:

2(sin x)^3 - sin x = 0

We'll factorize by sin x:

sin x[2(sin x)^2 - 1] = 0

We'll set each factor as 0.

sin x = 0

x = (-1)^k*arcsin 0 + kpi

x = kpi

2(sin x)^2 - 1 = 0

2(sin x)^2 = 1

(sin x)^2 = 1/2

sin x = +sqrt2/2

sin x = -sqrt2/2

x = (-1)^k*pi/4 + kpi

x = (-1)^(k+1)*pi/4 + kpi

**The solutions of the equation are: {kpi}U{(-1)^k*pi/4 + kpi}U{(-1)^(k+1)*pi/4 + kpi}.**