Question

sin x+ a sin(2x)+ sin(3x) = 0
x=?

Answer 1

$\sin \left(x\right)+\sin \left(2x\right)+\sin \left(3x\right)=0$

$\sin \left(3x\right)+\sin \left(x\right)+2\cos \left(x\right)\sin \left(x\right)=0$

$\sin \left(x\right)-\sin ^3\left(x\right)+2\cos \left(x\right)\sin \left(x\right)+3\cos ^2\left(x\right)\sin \left(x\right)=0$$\sin \left(x\right)\left(3\cos ^2\left(x\right)+2\cos \left(x\right)+1-\sin ^2\left(x\right)\right)=0$

$\sin \left(x\right)=0\quad \mathrm{sau}\quad \:3\cos ^2\left(x\right)+2\cos \left(x\right)+1-\sin ^2\left(x\right)=0$

$Pentru \sin \left(x\right)=0\\x=2\pi n,\:x=\pi +2\pi n$

$Pentru 3\cos ^2\left(x\right)+2\cos \left(x\right)+1-\sin ^2\left(x\right)=0$

$\cos ^2\left(x\right)+2\cos \left(x\right)+3\cos ^2\left(x\right)=0$

$2\cos \left(x\right)+4\cos ^2\left(x\right)=0$

$Fie \cos \left(x\right)=u$

$2u+4u^2=0$

Rezolvam ecuatia de gradul doi si avem

$u=0,\:u=-\frac{1}{2}$

$\cos \left(x\right)=0,\:\cos \left(x\right)=-\frac{1}{2}$

$x=\frac{\pi }{2}+2\pi n,\:x=\frac{3\pi }{2}+2\pi n,\:x=\frac{2\pi }{3}+2\pi n,\:x=\frac{4\pi }{3}+2\pi n$

Soluțiile sunt

$x=2\pi n,\:x=\pi +2\pi n,\:x=\frac{\pi }{2}+2\pi n,\:x=\frac{3\pi }{2}+2\pi n,\:x=\frac{2\pi }{3}+2\pi n,\:x=\frac{4\pi }{3}+2\pi n$

Answer 2

sin x +sin(2x) +sin(3x) =0 .

sin x +sin(3x) =2·sin(2x)·cos(-x) ,dar deoarece cos(x) este o functie para =>

cos(x)=cos(-x) => sin x +sin(3x) =2·sin(2x)·cos x =>

sin(2x) +2·sin(2x)·cos x =0 <=> sin(2x)·(1 +2·cos x)=0 =>

sin(2x)=0 => 2·sin x ·cos x=0 <=> sin x ·cos x=0 => sin x=0 => x∈{0;180;360} sau cos x=0 => x∈{90;270} .

sau 1 +2·cos x=0 <=> cos x=-1/2 => nu avem solutii in R .

Asadar x∈{0;90;180;270;360} <=> x∈{0;π/2;π;3π/2;2π) .