a-maths!~

2008-03-03 3:07 pm
解下列方程,其中0≤x≤2π。答案以π表示。




3(sec2x+cot2x)=13

回答 (1)

2008-03-04 12:49 am
✔ 最佳答案
We can use the following identity to simplify the question.
1 + tan^2 x = sec^2 x

3(sec^2 x + cot^2 x) = 13
3(1 + tan^2 x + cot^2 x) = 13
3tan^2 x + 3cot^2 x = 10
Multiply both sides by tan^2 x, we have
3tan^4 x + 3 = 10tan^2 x
3tan^4 x - 10tan^2 x + 3 = 0
(tan^2 x - 3)(3tan^2 x - 1) = 0
tan^2 x = 3 or 1/3
tan x = +- sqrt 3 or +- sqrt. 1/3
x = π/3, 2π/3, 4π/3, 5π/3 OR π/6, 5π/6, 7π/6, 11π/6


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