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Solve the equation over the interval [0,2pie), stating answers in radians

 Mar 10, 2019
 #1
avatar+111394 
+1

sin^2x - 2sin x + 1  = 0        factor as

 

(sin  x - 1) (sin x - 1) = 0

 

Either factor will give the same solution

 

sin x - 1  =   0

 

sin x = 1

 

arcsin (1)  = x  =  pi/ 2

 

 

cool  cool cool

 Mar 10, 2019
 #2
avatar+315 
+1

Thank you! smiley

Cupcake  Mar 10, 2019
 #3
avatar+111394 
0

No prob  !!!

 

cool cool cool

CPhill  Mar 10, 2019

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