+0  
 
+1
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avatar+295 

solve the equation on the interval [0,2π]

\(\sqrt{2}cos(2\Theta)=1 \)

 

A. None of these

B. π/8, 7π/8, 9π/8, 15π/8

C. π/2, π, 3π/2, 2π

D. 5π/8, 7π/8, 13π/8, 15π/8

E. π/4, π/2, 3π/4, π

Deathstroke_rule  Apr 25, 2017
 #1
avatar+89775 
+1

 

√2 cos(2 θ )   =  1     divide both sides by  √2

 

cos (2 θ )  =   1 / √2

 

Let  2 θ   = x

 

cos ( x)  =  1 / √2       and this happens at x  =   pi/4 , 7pi/4 , 9pi/4 , 15pi/4

 

So 

 

2  θ     =  pi/4,  7pi/4 , 9pi/4, 15pi/4         divide by 2

 

 θ   =    pi / 8 , 7pi/ 8, 9pi / 8, 15pi / 8       answer  "B"

 

 

cool cool cool

CPhill  Apr 25, 2017

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