How do you solve $sin x + sin 2 x + sin 3 x + sin 4 x = 0$ $Sin x + Sin 2x + Sin 3x + Sin 4x = 0$ ?

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How do you solve $sin x + sin 2 x + sin 3 x + sin 4 x = 0$ $Sin x + Sin 2x + Sin 3x + Sin 4x = 0$ ?

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Answer 1 · 2015-05-10T16:01:06

In this way, using the sum-to-product formula for sinus and cosine:

$sin α + sin β = 2 sin (\frac{α + β}{2}) cos (\frac{α - β}{2})$ $sinalpha+sinbeta=2sin((alpha+beta)/2)cos((alpha-beta)/2)$

$cos α + cos β = 2 cos (\frac{α + β}{2}) cos (\frac{α - β}{2})$ $cosalpha+cosbeta=2cos((alpha+beta)/2)cos((alpha-beta)/2)$ .

So:

$(sin 4 x + sin x) + (sin 3 x + sin 2 x) = 0 \Rightarrow$ $(sin4x+sinx)+(sin3x+sin2x)=0rArr$

$2 sin (\frac{4 x + x}{2}) cos (\frac{4 x - x}{2}) +$ $2sin((4x+x)/2)cos((4x-x)/2)+$

$+ 2 sin (\frac{3 x + 2 x}{2}) cos (\frac{3 x - 2 x}{2}) = 0$ $+2sin((3x+2x)/2)cos((3x-2x)/2)=0$

$2 sin (\frac{5}{2} x) cos (\frac{3}{2} x) + 2 sin (\frac{5}{2} x) cos (\frac{1}{2} x) = 0$ $2sin(5/2x)cos(3/2x)+2sin(5/2x)cos(1/2x)=0$

$2 sin (\frac{5}{2} x) [cos (\frac{3}{2} x) + cos (\frac{1}{2} x)] = 0$ $2sin(5/2x)[cos(3/2x)+cos(1/2x)]=0$

$2 sin (\frac{5}{2} x) 2 cos (\frac{\frac{3}{2} x + \frac{1}{2} x}{2}) cos (\frac{\frac{3}{2} x - \frac{1}{2} x}{2}) = 0$ $2sin(5/2x)2cos((3/2x+1/2x)/2)cos((3/2x-1/2x)/2)=0$

$4 sin (\frac{5}{2} x) cos x cos (\frac{x}{2}) = 0$ $4sin(5/2x)cosxcos(x/2)=0$ .

Then:

$sin (\frac{5}{2} x) = 0 \Rightarrow \frac{5}{2} x = k π \Rightarrow x = \frac{2}{5} k π$ $sin(5/2x)=0rArr5/2x=kpirArrx=2/5kpi$ ,

$cos x = 0 \Rightarrow x = \frac{π}{2} + k π$ $cosx=0rArrx=pi/2+kpi$ ,

$cos (\frac{x}{2}) = 0 \Rightarrow \frac{x}{2} = \frac{π}{2} + k π \Rightarrow x = π + 2 k π$ $cos(x/2)=0rArrx/2=pi/2+kpirArrx=pi+2kpi$ .

Answer 2 · 2015-05-23T13:15:59

Reminder of Concept to solve trig equations:
To solve a complex trig equation, transform it into a few basic trig equations. Solving trig equations finally results in solving basic trig equations.
In this example of solving: f(x) = sin x + sin 2x + sin 3x + sin 4x = 0, use the trig identity of "sin a + sin b" to transform f(x) into a product of 3 basic trig equations like Massimiliano finely did:
$f (x) = 4 cos x . sin (5 \frac{x}{2}) . cos (\frac{x}{2}) = 0$ $f(x) =4cos x.sin (5x/2).cos (x/2) = 0$ .
Next, solve separately the 3 basic trig equations: cos x = 0, sin $(5 \frac{x}{2}) = 0, and cos (\frac{x}{2}) = 0$ $(5x/2) = 0, and cos (x/2) = 0$

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How do you solve $sin x + sin 2 x + sin 3 x + sin 4 x = 0$ $Sin x + Sin 2x + Sin 3x + Sin 4x = 0$ ?

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