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Find all solutions of the equation in the interval [0, 2π). (Enter your answers as a comma-separated list. If there is no solution, enter NO SOLUTION.)

8 sec^2 x + 4 tan^2 x − 12 = 0

 Apr 7, 2016
 #1
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8 sec^2 x + 4 tan^2 x − 12 = 0     →     sec^2x = tan^2x + 1

 

8 [ tan^2x + 1] + 4tan^2x - 12  = 0

 

8tan^2x + 8 + 4tan^2x - 12  = 0

 

12tan^2x - 4   = 0     divide through by 4

 

3tan^2x - 1  = 0

 

3tan^2x  =  1  

 

tan^2x  = 1/3      take the square root of both sides

 

tanx = ±1 / √3

 

And on the given interval, this happens at     pi/6 , 5pi/6, 7pi/6, 11pi/6

 

 

cool cool cool

 Apr 7, 2016

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