SOLUTION: Find to the nearest degree, all values of theta in the interval 0 degrees is < or equal to theta which is < 180 degrees, that satisfy the equation:
3tan^2 theta+(1/cot theta)= 2
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-> SOLUTION: Find to the nearest degree, all values of theta in the interval 0 degrees is < or equal to theta which is < 180 degrees, that satisfy the equation:
3tan^2 theta+(1/cot theta)= 2
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Question 611001: Find to the nearest degree, all values of theta in the interval 0 degrees is < or equal to theta which is < 180 degrees, that satisfy the equation:
3tan^2 theta+(1/cot theta)= 2 Answer by lwsshak3(11628) (Show Source):
You can put this solution on YOUR website! Find to the nearest degree, all values of theta in the interval 0 degrees is < or equal to theta which is < 180 degrees, that satisfy the equation:
3tan^2 theta+(1/cot theta)= 2
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3tan^2x+(1/cotx)=2
3tan^2x+tanx-2=0
(3tanx-2)(tanx+1)=0
..
3tanx-2=0
tanx=2/3
x≈37º and 214º (in quadrants I and III where tan>0))
..
tanx+1=0
tanx=-1
x=135º and 315º (in quadrants II and IVI where tan<0))
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