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人気のある三角関数 >

tan^2(x)-1=0,0<= x<= 2pi

解

tan^{2} (x) - 1 = 0, 0 \leq x \leq 2 π

解

x = \frac{π}{4}, x = \frac{5 π}{4}, x = \frac{3 π}{4}, x = \frac{7 π}{4}

+1

度

x = 4 5^{\circ}, x = 22 5^{\circ}, x = 13 5^{\circ}, x = 31 5^{\circ}

解答ステップ

tan^{2} (x) - 1 = 0, 0 \leq x \leq 2 π

置換で解く

tan^{2} (x) - 1 = 0

仮定:

tan (x) = 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 = tan (x)

tan (x) = 1, tan (x) = - 1

tan (x) = 1, tan (x) = - 1

tan (x) = 1, 0 \leq x \leq 2 π : x = \frac{π}{4}, x = \frac{5 π}{4}

tan (x) = 1, 0 \leq x \leq 2 π

以下の一般解

tan (x) = 1

tan (x)

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

範囲の解答

0 \leq x \leq 2 π

x = \frac{π}{4}, x = \frac{5 π}{4}

tan (x) = - 1, 0 \leq x \leq 2 π : x = \frac{3 π}{4}, x = \frac{7 π}{4}

tan (x) = - 1, 0 \leq x \leq 2 π

以下の一般解

tan (x) = - 1

tan (x)

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

範囲の解答

0 \leq x \leq 2 π

x = \frac{3 π}{4}, x = \frac{7 π}{4}

すべての解を組み合わせる

x = \frac{π}{4}, x = \frac{5 π}{4}, x = \frac{3 π}{4}, x = \frac{7 π}{4}

グラフ

人気の例

sin^{2} (θ) = \frac{1}{4}

(tan(θ)+1)(sec(θ)+1)=0

(tan (θ) + 1) (sec (θ) + 1) = 0

2sin(θ)-tan^2(θ)=0

2 sin (θ) - tan^{2} (θ) = 0

sin(θ)-sqrt(3)cos(θ)=1

sin (θ) - 3 cos (θ) = 1

cos (a) = \frac{1}{2}