2017-02-10 05:06:18nzphddr
解下列問題
標題:
解下列問題
發問:
範圍: 0° 少過or等如 x 少過or等如 360° 3) 3 tan^2 x = 1 16) 2 sinx cosx - sinx + 2cosx - 1 = 0 17) 2開方3 cos^2x + cosx - 2開方3 = 0 18) 3 cot^2x + ( 3 + 開方3 ) cotx + 開方3 = 0 19) 3( sec^2x + xot^2x) = 13 更新: 19) 3( sec^2x + cot^2x) = 13 更新 2: 19) 3( sec^2 x + cot^2 x) = 13 18) 3 cot^2 x + ( 3 + 開方3 ) cotx + 開方3 = 0 17) 2開方3 cos^2 x + cosx - 2開方3 = 0
最佳解答:
3) 3 tan2 x = 1 tan2x = 1/3 tan x = ± 1/√3 x = 30° or 150° or 210° or 330° 16) 2 sinx cosx - sinx + 2cosx - 1 = 0 sin x(2cos x - 1) + 2cos x - 1 = 0 (2cos x - 1)(sin x + 1) = 0 cos x = 1/2 or sin x = -1 x = 60° or 300° or 270° x = 60° or 270° or 300° 17) 2√3 cos2x + cosx - 2√3 = 0 (2cos x - √3)(√3 cos x + 2) = 0 cos x = √3/2 or cos x = -2/√3 (rej) x = 30° or 330° 18) 3 cot2x + ( 3 + √3 ) cotx + √3 = 0 ( √3 cot x +1 )(√3cot x + √3) = 0 cot x = -1/√3 or cot x = -1 tan x = -√3 or tan x = -1 x = 120° or 300° or 135° or 315° x = 120° or 135° or 300° or 315° 19) 3( sec2x + cot2x) = 13 sec2x + cot2x = 13/3 1 + tan2x + cot2x = 13/3 tan2x - 10/3 + cot2x = 0 tan^4 x - (10/3)tan2x + 1 = 0 3tan^4 x - 10tan2x + 3 = 0 (tan2x - 3)(3tan2x - 1) = 0 tan2x = 3 or tan2x = 1/3 tan x = ±√3 or tan x = ± 1/√3 x = 60° or 120° or 240° or 300° or 30° or 150° or 210° or 330° x = 30° or 60° or 120° or 150° or 210° or 240° or 300° or 330°
其他解答:
解下列問題
發問:
範圍: 0° 少過or等如 x 少過or等如 360° 3) 3 tan^2 x = 1 16) 2 sinx cosx - sinx + 2cosx - 1 = 0 17) 2開方3 cos^2x + cosx - 2開方3 = 0 18) 3 cot^2x + ( 3 + 開方3 ) cotx + 開方3 = 0 19) 3( sec^2x + xot^2x) = 13 更新: 19) 3( sec^2x + cot^2x) = 13 更新 2: 19) 3( sec^2 x + cot^2 x) = 13 18) 3 cot^2 x + ( 3 + 開方3 ) cotx + 開方3 = 0 17) 2開方3 cos^2 x + cosx - 2開方3 = 0
最佳解答:
3) 3 tan2 x = 1 tan2x = 1/3 tan x = ± 1/√3 x = 30° or 150° or 210° or 330° 16) 2 sinx cosx - sinx + 2cosx - 1 = 0 sin x(2cos x - 1) + 2cos x - 1 = 0 (2cos x - 1)(sin x + 1) = 0 cos x = 1/2 or sin x = -1 x = 60° or 300° or 270° x = 60° or 270° or 300° 17) 2√3 cos2x + cosx - 2√3 = 0 (2cos x - √3)(√3 cos x + 2) = 0 cos x = √3/2 or cos x = -2/√3 (rej) x = 30° or 330° 18) 3 cot2x + ( 3 + √3 ) cotx + √3 = 0 ( √3 cot x +1 )(√3cot x + √3) = 0 cot x = -1/√3 or cot x = -1 tan x = -√3 or tan x = -1 x = 120° or 300° or 135° or 315° x = 120° or 135° or 300° or 315° 19) 3( sec2x + cot2x) = 13 sec2x + cot2x = 13/3 1 + tan2x + cot2x = 13/3 tan2x - 10/3 + cot2x = 0 tan^4 x - (10/3)tan2x + 1 = 0 3tan^4 x - 10tan2x + 3 = 0 (tan2x - 3)(3tan2x - 1) = 0 tan2x = 3 or tan2x = 1/3 tan x = ±√3 or tan x = ± 1/√3 x = 60° or 120° or 240° or 300° or 30° or 150° or 210° or 330° x = 30° or 60° or 120° or 150° or 210° or 240° or 300° or 330°
其他解答:
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