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How would I do this?

solve the following on the domain [0,2π]

(cos x + 1)(√2 cos x - 1) = 0

 Jan 20, 2016
 #1
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(cos x + 1)(√2 cos x - 1) = 0      set each factor to 0

 

cos x + 1 = 0       add 1 to  both sides

 

cos x  = -1

 

And this happens at  pi  on  [0, 2pi]  

 

And

 

√2 cos x - 1  = 0      add 1 to both sides

 

√2 cos x  = 1          divide both sides by √2

 

cos x =    1 / √2

 

And this happens at   pi/4    and 7pi/4    on [0, 2pi]

 

 

cool cool cool

 Jan 20, 2016

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