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avatar+90 

Hey, could you help me with this question???

 

Solve the equation for the solution in the indicated interval.

 

\(\sqrt[2]{3}sec3\theta-2=0 \)    [0°,720°]  \(\leftarrow\)Interval

 

also, could you explain steps??

 

Thanks in advance!!

dom6547  Nov 30, 2017
edited by dom6547  Nov 30, 2017
 #1
avatar+89874 
+2

√3sec (3θ) -  2  = 0   add 2 to each side

 

√3sec (3θ)  =   2        divide both sides by  √3

 

sec (3θ )  =  2  / √3

 

Let q  =  3θ

 

sec (q)   = 2  / √3

 

And this happens when cos (q)  = √3 / 2    .....so....

 

q  =  30°,  330°, 390°, 690°, 750°, 1050° , 1110°, 1410°, 1470°, 1770°, 1830°, 2130° 

 

So  3θ  =   30°,  330°, 390°, 690°, 750°, 1050°, 1110°, 1410°, 1470°, 1770°, 1830°, 2130° 

 

Divide each side by 3

 

θ  =  10°, 110°, 130°, 230°, 250°, 350°, 370°, 470°, 490°, 590°, 610°, 710°

 

Here's the graph :  https://www.desmos.com/calculator/yj6jsij1kc

 

 

cool cool cool

CPhill  Nov 30, 2017
 #2
avatar+90 
+2

THANK YOU MUCH!!!

dom6547  Nov 30, 2017

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