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in the interval 0 'less than or equal to'  x  'less that or equal to'   pi

find the values that satisfy the equation:

cos 2x - sin x = 0

 Aug 19, 2016
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cos 2x - sin x = 0      note:  cos 2x =  1 - 2sin^2x

 

1 - 2sin^2x - sinx    =   0      multiply through by -1

 

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

 

(2sinx - 1) (sinx + 1)  =  0      set each factor to 0

 

2sinx  - 1  = 0    add 1 to both sides

 

2sinx  = 1      divide both sides by  2

 

sinx   = 1/2       and this happens at   pi/6  and 5pi/6   on the requested interval

 

And

 

sin x + 1 = 0    subtract 1 from both sides

 

sin x = -1       this happens at 3pi/2, which is outside the requested interval

 

 

 

cool cool cool

 Aug 20, 2016

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