sec^2 x + tan^2 x=3
in the intercal [0,2pi)
2007-12-28 09:31:12 · 7 answers · asked by em.four 2 in Science & Mathematics ➔ Mathematics
sec²x + tan²x = 3
Remember that sec²x = 1 + tan²x
1 + tan²x + tan²x = 3
2tan²x = 2
tan²x = 1
tan(x) = ±1
x = {π/4, 3π/4, 5π/4, 7π/4}
2007-12-28 09:36:18 · answer #1 · answered by gudspeling 7 · 0⤊ 0⤋
Substitute 1 + tan^2 x for the sec^2 x:
1 + tan^2 x + tan^2 x = 3
1 + 2 tan^2 x = 3
2 tan^2 x = 2
tan^2 x = 1
tan x = +/- 1
This would occur at pi/4, 3pi/4, 5pi/4, and 7pi/4.
that's it
2007-12-28 09:38:40 · answer #2 · answered by Marley K 7 · 0⤊ 0⤋
sec^2 x + tan^2 x=3
=> 1 + tan^2 x + tan^2 x = 3
=> 2 tan^2 x = 2
=> tan^2 x = 1
=> tan x = ± 1 = tan (± π/4)
=> x = kπ ± π/4, where k ∈ Z.
As x ∈ [0,2pi),
x = π/4, 3π/4, 5π/4 and 7π/4.
2007-12-28 09:36:32 · answer #3 · answered by Madhukar 7 · 0⤊ 0⤋
1 + tan^x + tan^2 x = 3
2tan^2 x = 2
tan^2 x = 1
tan x = +/-1
x = pi/4, 3pi/4, 5pi/4, or 7pi/4
2007-12-28 09:37:01 · answer #4 · answered by norman 7 · 0⤊ 0⤋
tan ² x + 1 + tan ² x = 3
2 tan ² x = 2
tan ² x = 1
tan x = ± 1
x = π/4 , 3π/4 , 5π/4 , 7π/4
2007-12-28 21:20:43 · answer #5 · answered by Como 7 · 4⤊ 0⤋
sec^2x = tan^2x + 1
so
tan^2x + 1 +tan^2x = 3
2tan^2x = 2
tan^2x = 1
tanx = 1 , x = pi / 4 , 5pi/4,
and tanx = - 1 , x= - pi / 4 , - 3pi/4 , 3pi/4,7pi/4
so
solution is
{x = - 3pi/ 4 ,- pi /4 , pi/4, 3pi/4 , 5pi/4 , 7pi / 4 }
2007-12-28 09:49:56 · answer #6 · answered by LE THANH TAM 5 · 0⤊ 2⤋
1+sin^2(x) = 3*cos^2(x)
2 = 4cos^2(x)
cos(x) = 1/root(2)
x = pi/4
2007-12-28 09:51:49 · answer #7 · answered by Nur S 4 · 0⤊ 2⤋