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What are the solutions to the trigonometric equation on the interval [0,2π) ?

2sin2x=sinx

Select all correct solutions.

0

π/6

π/3

π/2

2π/3

5π/6

π

7π/6

4π/3

3π/2

5π/3

11π/6

 Apr 30, 2020
 #1
avatar+111438 
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I think this is supposed to be

 

2sin^2 x  = sin x

 

2*sin^2 x - sin x  =  0              factor out sin x

 

sin x ( 2sin x - 1)  =  0

 

sin x  = 0       and this occurs when x = 0   and x = pi

 

And

 

2sin x - 1  = 0

2sin x  = 1

sin x  =1/2    and this occurs when x = pi/6  and x  = 5pi/6

 

 

cool cool cool

 Apr 30, 2020

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