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Assuming your equation is: 6(tanx)2-17tanx + 7 = 0
The easiest way to do it is use a substitution. Let s=tanx, then substitute s for tanx in the original equation to get the following:
6s2-17s+7=0
Using the quadratic equation solve for s:
s = {-(-17) +- SQRT [(-17)2 - 4*(6)*7)]}/(2*6)
s = [17 + SQRT (121)]/12 & s= [17 - SQRT(121)]/12
s = 28/12 or 7/3 & s = 6/12 or 1/2
Now substitute tanx back for s to get
tan x = 7/3 & tan x = 1/2
x = tan-1(7/3) or approx. 66.8o & x= tan-1(1/2) or approx. 26.6o
In radians it would be 1.166 & 0.4636
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∙ 12y ago
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