solve this problem! first true answer gets best answer!?

Question

sin x + sin 2x + sin 3x = 0 , 0<= x =< 2π find x ( more than 1 answer)

Answer 1

sin x + sin 3x + sin 2x =0
2 sin 2x * cos x + sin 2x =0
sin 2x (2 cos x + 1) = 0

=> sin 2x = 0 or cos x = -1/2

sin2x=0 => 2x = 0, π , 2π, 3π, 4π
=> x= 0, π/2, π, 3π/2, 2π

cos x = -1/2 => x= 2π/3, 4π/3

Solution is 0, π/2, 2π/3, π, 4π/3, 3π/2, 2π

Answer 2

You can get the answer by using Transformations.

Apply the formula, sin c + sin d = 2 Sin(c+d)/2 * Cos(c-d)/2 to the first two terms (sin x + sin 2x). And use Sin A = 2 SinA/2 CosA/2 for the third expresion (sin 3x ).

This gives,

2 Sin(3/2x) Cos(x/2) + 2 Sin(3x/2) Cos(3x/2) = 0

Cancelling 2 and taking sin(3x/2) as common.

Sin(3x/2) { Cos(x/2) + Cos(3x/2) } = 0

Apply the formula, cos c + cos d = 2 cos(c+d)/2 * Cos(c-d)/2 in the above expression.

Gives,
Sin(3x/2) { 2 Cosx Cos(x/2) } = 0

This gives that Sin(3x/2) = 0 , Cosx = 0 , Cos(x/2) = 0.

Taking, Sin(3x/2) = 0
gives, 3x/2 = 0 , π As sin R = 0 when R = 0,π in the interval of 0<= R =< 2π
gives, x = 0 , 2π/3

Now taking, Cosx = 0
gives, x = π/2 , 3π/2.

Now taking, Cos(x/2) = 0
gives, (x/2) = π/2 , 3π/2. As Cos R = 0 when R = π/2 , 3π/2 in the interval of 0<= R =< 2π
gives, x = π , 3π

So the solutions for x that we got are 0 , 2π/3 , π/2 , 3π/2 , π , 3π.

In the question you gave that x should belong to [0,2π].
So the values of x are 0 , 2π/3 , π/2 , π , 3π/2.

Hence this is the solution.

Answer 3

sin(x) + sin(2x) + sin(3x) = 0
sin(x) + (2sinxcosx) + (-sinx^3 + 3cosx^2 * sinx) = 0
sin(x) + 2sinxcosx - sinx^3 + 3sinxcosx^2 = 0
(sin(x))(1 + 2cosx - sinx^2 + 3cosx^2) = 0
(sin(x))(3cosx^2 + 2cosx - sinx^2 + 1) = 0
(sin(x))(3cosx^2 + 2cosx + 1 - sinx^2) = 0
(sin(x))(3cosx^2 + 2cosx + cosx^2) = 0
(sin(x))(4cosx^2 + 2cosx) = 0
(sinx)(2cosx)(2cosx + 1) = 0

sinx = 0
x = 0° or 180°

2cosx = 0
cosx = 0
x = 90° or 270°

2cosx + 1 = 0
2cosx = -1
cosx = (-1/2)
x = 120° or 240°

x = 0°, 90°, 120°, 180°, 240°, or 270°

in radians that would be

x = 0, (pi/2), (2pi/3), pi, (4pi/3), (3pi/2), or 2pi

Answer 4

apply the formula sin A +sin B = 2sin((A+B)/2)cos((A-B)/2)

=> 2sin3x/2 cosx/2 + sin3x = 0

but sin3x = 2sin3x/2 cos3x/2


=> 2sin3x/2 (cosx/2 + cos3x/2) = 0

cos A + cos B = 2cos((A+B)/2)cos((A-B)/2)

=> 4sin3x/2 cos2x cosx =0

=> sin3x/2 = 0 or cos2x = 0 or cos x=0

=> x=0 , 120 or x=180 or x=90

the possible values are x=0,90,120,180

Answer 5

Romans 6:23 For the wages of sin is death; but the gift of God is eternal life through Jesus Christ our Lord.
Romans 8:2 For the law of the Spirit of life in Christ Jesus hath made me free from the law of sin and death.
Hebrews 12:1 Wherefore seeing we also are compassed about with so great a cloud of witnesses, let us lay aside every weight, and the sin which doth so easily beset us, and let us run with patience the race that is set before us,
1 John 1:7 But if we walk in the light, as he is in the light, we have fellowship one with another, and the blood of Jesus Christ his Son cleanseth us from all sin.

Answer 6

won't be able to remedy it purely simplify by using creating denominators into: (x+3)(x+2) and (x+7)(x+2) and (x+3)(x+7) Then split utilising partial fractions. carry jointly numerators 2gether. answer my parenting Q on toddler-proofing mouths. a lot more effective not easy.

Answer 7

sinx+sin2x+sin3x=0
sinx+sin2x=2sin3x/2cosx
and sin3x=2sin3x/2cos3x/2 and so
sinx+sin2x+sin3x
=2sin3x/2(cosxcos3x/2)
if sinx+sin2x+sin3x=0
sin3x/2=0 or cosx=0 or cos3x/2=0
so using the general solution
for sinx=o or cosx=0
sin3x/2=0 will mean 3x/2=n*pi or x=2n*pi/3
cosx=0 will mean x=(2n+1)*pi/2 and
cos3x/2=0 will mean 3x/2=(2n+1)*pi/2
=>x=(2n+1)*pi/3
when x is bet. 0 and pi
the values can be
0,pi/3,pi/2,2*pi/3,4*pi/3,3*pi/2,pi,5*pi/3
or in degrees it shall be
0deg,60deg,90deg,120deg,180deg,
240deg,270deg,300deg

Answer 8

one of the ans is x=0

Answer 9

Plug it in a graphing calculator

Answer 10

sin is sin and hell is the answer. I am good.