+0  
 
0
123
1
avatar+4 

Find x such that sin2x=sinx & 0<=x<=2pi.

 Feb 15, 2022
 #1
avatar+124594 
+1

sin 2x  =  sin x      (using an identity on the left and simplifying)

 

2sinxcosx  - sin x  =  0            factor out sin x

 

sin x ( 2cos x  - 1)  =   0

 

We have two equations

 

sin x = 0    and this happens  at 0 , pi  and 2pi

 

2cos x - 1 =  0

2cos x  =  1

cos x =1/2    and this happens at pi/3  and  5pi/ 3

 

So the solutions are   0 , pi/3 , pi, 5pi/3  and 2 pi

 

 

cool cool cool

 Feb 15, 2022

20 Online Users