+0  
 
0
768
5
avatar

sinx-2sinxcosx=0

 Feb 6, 2017
 #1
avatar+37084 
0

re-arrange to get

2 sinx cos x = sinx     divide both sides by sinx

2 cosx = 1

cosx = 1/2

x = arcos(1/2) = 60 degrees (pi/3)    or   - 60 degrees  (- pi/3)

 Feb 6, 2017
 #2
avatar+129839 
0

Gotta' be careful here, EP.....dividing out sin x is like throwing away a polynomial root....

 

sin x  - 2sin x cos x  = 0

 

sin x ( 1 - 2cosx)  = 0

 

So

 

sin x  = 0    at    0° + n*180°   where n is an integer    .....or

 

1 - 2cosx  = 0

 

1 = 2cos x

 

1/2  = cos x        at    60° + n360°    and at  300° + n360°

 

 

cool cool cool

 Feb 6, 2017
 #4
avatar+37084 
0

Thanx for the heads up Chris.....I did check the results to be sure before I posted (this time)  !

indecision

ElectricPavlov  Feb 6, 2017
 #5
avatar+37084 
0

.....but I didn't get all of the possible answers.....just like I didn't finish that last post before hitting 'enter'     LOL   indecisionindecision

ElectricPavlov  Feb 6, 2017
 #3
avatar+26387 
+5

sinx-2sinxcosx=0

 

Formula:

\(\begin{array}{|rcll|} \hline 2\ sin(x) \cos(x) &=& \sin(2x)\\ \hline \end{array}\)

 

 

So:

\(\begin{array}{|rcll|} \hline 0 &=& \sin(x) - 2\cdot sin(x) \cos(x) \quad & | \quad 2\cdot sin(x) \cos(x) = \sin(2x) \\ 0 &=& \sin(x) - \sin(2x) \quad & | \quad + \sin(2x) \\ \sin(2x) &=& \sin(x) \\ \hline \end{array} \)

 

First result:

\(\begin{array}{|rcll|} \hline \sin(2x) &=& \sin(x) \quad & | \quad \arcsin() \text{ both sides } \\ 2x &=& x \quad & | \quad -x \\ 2x-x &=& 0 \\ \mathbf{ x } & \mathbf{=} & \mathbf{ 0 \pm 360^{\circ} \cdot n } \quad & | \quad n \in N \\ \hline \end{array}\)

 

Second result:

\(\begin{array}{|rcll|} \hline \sin(2x) &=& \sin(x) \quad & | \quad \sin(x) = \sin(180^{\circ}-x) \\ \sin(2x) &=& \sin(180^{\circ}-x) \quad & | \quad \arcsin() \text{ both sides } \\ 2x &=& 180^{\circ}-x \quad & | \quad +x \\ 2x+x &=& 180^{\circ} \\ 3x &=& 180^{\circ} \quad & | \quad : 3 \\ x &=& 60^{\circ} \\ \mathbf{ x } & \mathbf{=} & \mathbf{ 60^{\circ} \pm 360^{\circ} \cdot n } \quad & | \quad n \in N \\ \hline \end{array}\)

 

Third result:

\(\begin{array}{|rcll|} \hline \sin(2x) &=& \sin(x) \quad & | \quad -\sin(x) = \sin(180^{\circ}+x) \\ \sin(2x) &=& -\sin(180^{\circ}+x) \\ \sin(2x) &=& \sin(-180^{\circ}-x) \quad & | \quad \arcsin() \text{ both sides } \\ 2x &=& -180^{\circ}-x \quad & | \quad +x \\ 2x+x &=& -180^{\circ} \\ 3x &=& -180^{\circ} \quad & | \quad : 3 \\ x &=& -60^{\circ} \\ \mathbf{ x } & \mathbf{=} & \mathbf{ -60^{\circ} \pm 360^{\circ} \cdot n } \quad & | \quad n \in N \\ \hline \end{array} \)

 

 

laugh

 Feb 6, 2017

6 Online Users

avatar
avatar