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2sec^2x -3tanx =5 for [0,2pi)
7 years ago

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bobpursley
tan2(x) + 1 = sec2(x) identity
2(tan^2(x)+1)-3tanx=5
put that in standard form
au^2+bu+c=0 and solve for tan x
7 years ago
tom
and what are the answers
7 years ago

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if f(x)=3tanx +2 Solve: sec(x)^2= 3tanx +1 *** It is suppose to be the sqaure root of 3 but i could find th... f(x)= 2sec(x-3.14/4)-1 tan3x=(3tanx-tan^3x)/(1-3tan^2x) solve: tanx+tan2x+√3tanx tan2x=√3 (0<=x<=360) 2sec^2x - 3tanx - 5 = 0 I got x= 1.24 radians x= 4.38 radians x= -0.74 radians x= -3.88 rad... 2sec^2x -3tanx =5 for [0,2pi) Please solve for the values of x for me. 2sec^2x - 3tanx - 5 = 0 over the interval 0 to 2pi x= 1.14, 4.28, 5.68, and 2.54 right? t=2sec, s=2m,v=? Differentiate x-3+3tanx
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