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Determine the exact value of x. cos2x-sinx=0

 Feb 5, 2019
 #1
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 cos2x -  sinx = 0

 

(1 - 2sin^2 x) - sin x  = 0             multiply through by -1

 

2sin^2 (x)  +  sin x - 1   =  0       factor

 

(2sin x - 1) (sin x + 1) = 0

 

Set each factor to 0   and solve for  x

 

2sinx - 1 = 0

2sin x = 1

sin x = 1/2

 

And this happens at  x =   pi/6 + 2pi*n       and     x =   5pi/6 + 2pi*n       where n is an integer

 

Also

 

sinx + 1  = 0

sin x = -1

 

And this happens at     x  =   3pi/2 + 2pi*n    where n is an integer  

 

 

 

cool cool cool

 Feb 5, 2019

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