ja

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

人気のある三角関数 >

sec^2(θ)-1=0,0<= θ<= 2pi

解

sec^{2} (θ) - 1 = 0, 0 \leq θ \leq 2 π

解

θ = 0, θ = 2 π, θ = π

+1

度

θ = 0^{\circ}, θ = 36 0^{\circ}, θ = 18 0^{\circ}

解答ステップ

sec^{2} (θ) - 1 = 0, 0 \leq θ \leq 2 π

置換で解く

sec^{2} (θ) - 1 = 0

仮定:

sec (θ) = u

u^{2} - 1 = 0

u^{2} - 1 = 0 : u = 1, u = - 1

u^{2} - 1 = 0

1

を右側に移動します

u^{2} - 1 = 0

両辺に

1

を足す

u^{2} - 1 + 1 = 0 + 1

簡素化

u^{2} = 1

u^{2} = 1

x^{2} = f (a)

の場合, 解は

x = f (a), - f (a)

u = 1, u = - 1

1 = 1

1

規則を適用

1 = 1

= 1

- 1 = - 1

- 1

規則を適用

1 = 1

= - 1

u = 1, u = - 1

代用を戻す

u = sec (θ)

sec (θ) = 1, sec (θ) = - 1

sec (θ) = 1, sec (θ) = - 1

sec (θ) = 1, 0 \leq θ \leq 2 π : θ = 0, θ = 2 π

sec (θ) = 1, 0 \leq θ \leq 2 π

以下の一般解

sec (θ) = 1

sec (x)

2 πn

循環を含む周期性テーブル :

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sec (x) 1 \frac{2 3}{3} 22 Undefined - 2 - 2 - \frac{2 3}{3} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sec (x) - 1 - \frac{2 3}{3} - 2 - 2 Undefined 22 \frac{2 3}{3}

θ = 0 + 2 πn

θ = 0 + 2 πn

解く

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn

範囲の解答

0 \leq θ \leq 2 π

θ = 0, θ = 2 π

sec (θ) = - 1, 0 \leq θ \leq 2 π : θ = π

sec (θ) = - 1, 0 \leq θ \leq 2 π

以下の一般解

sec (θ) = - 1

sec (x)

2 πn

循環を含む周期性テーブル :

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sec (x) 1 \frac{2 3}{3} 22 Undefined - 2 - 2 - \frac{2 3}{3} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sec (x) - 1 - \frac{2 3}{3} - 2 - 2 Undefined 22 \frac{2 3}{3}

θ = π + 2 πn

θ = π + 2 πn

範囲の解答

0 \leq θ \leq 2 π

θ = π

すべての解を組み合わせる

θ = 0, θ = 2 π, θ = π

グラフ

人気の例

2cos^2(θ)-3cos(θ)=0,(3pi)/2 <θ<2pi

2 cos^{2} (θ) - 3 cos (θ) = 0, \frac{3 π}{2} < θ < 2 π

cos (θ) = \frac{9}{17}

-1=(tan(x)-(3/4))/(1+tan(x)(3/4))

- 1 = \frac{tan ( x ) - ( \frac{3}{4} )}{1 + tan ( x ) ( \frac{3}{4} )}

cos (x) = - \frac{15}{13}

arctan((20)/x)-arctan((400)/x)=-9.72

arctan (\frac{20}{x}) - arctan (\frac{400}{x}) = - 9.72