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Help #50,51

 May 7, 2018
 #1
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50.  2sinθcosθ + √2 cos θ   = 0          factor out cosθ

 

cosθ (2sin θ + √2 )  = 0

 

So...either  cosθ  =  0      and this happpens at  pi/2  and  3pi/2     ......or....

 

2sin θ + √2   = 0       subtract √2  from both sides

 

2 sinθ  =  -√2    divide both sides by  2

 

sinθ  =   -√2 / 2     and this happens a 5pi/4    and at 7pi/4

 

So...the answers are    pi/2 , 3pi/2./  5pi/4 and  7pi/4   ⇒   the third answer

 

 

51.     √(7x) [ √x - 7√7  ] =

 

 √(7) x- 49√x

 

 

cool cool cool

 May 7, 2018

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