Respuesta :

prosborough198p2ub41
Using trig identity

sin(a)+sin(b)=2sin(a+b/2)*cos(a-b/2)
=2sin(2x)*cos(x)

Using your equation;

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

That now equals using sin(2x) double angle;

[2sin(2x)]*[2cos(x)+1]=0

We are left with a basic expression of A*B=0
Therefore now we know one or both expressions must equal zero to have the equation equal zero. We will solve for zero on each separately

If you don’t have it memorized yet(AND YOU DEFINITELY SHOULD DO THAT!!!) go ahead and use the unit circle to find where values of sin(a)=0 we find that is true when a=0,pi,2pi

sin(2x)=0

2x=0 therefore x=0
2x=pi therefore x=pi/2
2x=2pi therefore x=pi

General form x= k*pi/2
Where k=any even integer

Now to find when the other expression equals zero

2cos(x)+1=0 Using unit circle again for values when cos(a)=-1/2 remember NEGATIVE sign

cos(x)=-1/2 therefore;

cos(x)= +/- 3*pi/2, +2k*pi

Any questions please feel free to ask. One trick with more complicated trig expressions, they can always be simplified down to some basic trig expressions that can easily be solved. Practice practice practice to learn what formulas to use and when. Hope I helped and best of luck. Cheers!
gmany

Answer:

[tex]\large\boxed{x=k\pi\ \vee\ x=\dfrac{\pi}{2}+k\pi\ \vee\ x=\pm\dfrac{2\pi}{3}+2k\pi,\ k\in\mathbb{Z}}[/tex]

Step-by-step explanation:

[tex]\sin3x+\sin2x+\sin x=0\\\\\sin3x=\sin(2x+x)\\\text{use}\ \sin(a+b)=\sin a\cos b+\sin b\cos a\\\\=\sin2x\cos x+\sin x\cos2x\\\text{use}\ \sin2a=2\sin a\cos a\ \text{and}\ \cos2a=\cos^2x-\sin^2a\\\\=2\sin x\cos x\cos x+\sin x(\cos^2x-\sin^2x)\\=2\sin x\cos^2x+\sin x\cos^2x-\sin^3x\\=3\sin x\cos^2x-\sin^3x[/tex]

[tex]\text{come back to the equation:}\\\\3\sin x\cos^2x-\sin^3x+2\sin x\cos x+\sin x=0\qquad\text{distributive}\\\\\sin x(3\cos^2x-\sin^2x+2\cos x+1)=0\\\\\sin x(3\cos^2x+2\cos x+\underbrace{(1-\sin^2x)}_{=\cos^2x})=0\\\text{use}\ \sin^2x+\cos^2x=1\to\cos^2x=1-\sin^2x\\\\\sin x(3\cos^2x+2\cos x+\cos^2x)=0\\\\\sin x(4\cos^2x+2\cos x)=0\qquad\text{distributive}\\\\\sin x\cos x(4\cos x+2)=0\iff\sin x\cos x=0\ or\ 4\cos x+2=0\\\\\sin x\cos x=0\iff\sin x=0\ or\ \cos x=0[/tex]

[tex]\sin x=0\iff \boxed{x=k\pi},\ k\in\mathbb{Z}\\\\\cos x=0\iff \boxed{x=\dfrac{\pi}{2}+k\pi},\ k\in\mathbb{Z}\\\\4\cos x+2=0\qquad\text{subtract 2 from both sides}\\4\cos x=-2\qquad\text{divide both sides by 4}\\\cos x=-\dfrac{1}{2}\iff \boxed{x=\pm\dfrac{2\pi}{3}+2k\pi},\ k\in\mathbb{Z}[/tex]

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