# C2 questionWatch

#1
I need some help with this question

3tan^2 θ -1=0

for θ in the interval -pi<θ<pi

thanks
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12 years ago
#2
tan2x = 1/3
tanx = +/- 1/sqrt3
x = pi/6, -5pi/6, -pi/6, 5pi/6
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#3
how did you get?

x = pi/6, -5pi/6, -pi/6, 5pi/6
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12 years ago
#4
(Original post by domi1989)
how did you get?

x = pi/6, -5pi/6, -pi/6, 5pi/6
Use a 2 by 2 by 2 equilateral triangle - them half to get angles 30, 60 and 90. Frrom this tan 30 = 1/rt3

360/12 =30 therefore 2pi/12 = pi/6

Now use the CAST circle. Tangent is positive in 1st and 3rd quadrants and negative in 2nd and 4th.

SORRY for making the WRONG correction. I hope it is now ALL correct
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12 years ago
#5
(Original post by steve2005)
Use a 2 by 2 by 2 equilateral triangle - them half to get angles 30, 60 and 90. Frrom this tan 30 = 1/rt3

360/12 =30 therefore 2pi/12 = pi/6

Now use the CAST circle. Tangent is positive in 1st and 3rd quadrants and negative in 2nd and 4th.
Long time no see Steve, tan30=1/rt3 that angle in your diagram is for angle of 60degrees.
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12 years ago
#6
Hi typing error.... now corrected.
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