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人気のある三角関数 >

tan^2(x)sin(x)=sin(x),0<= x<= 2pi

解

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

解

x = 0, x = π, x = 2 π, x = \frac{3 π}{4}, x = \frac{7 π}{4}, x = \frac{π}{4}, x = \frac{5 π}{4}

+1

度

x = 0^{\circ}, x = 18 0^{\circ}, x = 36 0^{\circ}, x = 13 5^{\circ}, x = 31 5^{\circ}, x = 4 5^{\circ}, x = 22 5^{\circ}

解答ステップ

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

両辺から

sin (x)

を引く

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

因数

tan^{2} (x) sin (x) - sin (x) : sin (x) (tan (x) + 1) (tan (x) - 1)

tan^{2} (x) sin (x) - sin (x)

共通項をくくり出す

sin (x)

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

因数

tan^{2} (x) - 1 : (tan (x) + 1) (tan (x) - 1)

tan^{2} (x) - 1

1

を書き換え

1^{2}

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

2乗の差の公式を適用する:

x^{2} - y^{2} = (x + y) (x - y)

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

= (tan (x) + 1) (tan (x) - 1)

= sin (x) (tan (x) + 1) (tan (x) - 1)

sin (x) (tan (x) + 1) (tan (x) - 1) = 0

各部分を別個に解く

sin (x) = 0 or tan (x) + 1 = 0 or tan (x) - 1 = 0

sin (x) = 0, 0 \leq x \leq 2 π : x = 0, x = π, x = 2 π

sin (x) = 0, 0 \leq x \leq 2 π

以下の一般解

sin (x) = 0

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

解く

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

範囲の解答

0 \leq x \leq 2 π

x = 0, x = π, x = 2 π

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

tan (x) + 1 = 0, 0 \leq x \leq 2 π

1

を右側に移動します

tan (x) + 1 = 0

両辺から

1

を引く

tan (x) + 1 - 1 = 0 - 1

簡素化

tan (x) = - 1

tan (x) = - 1

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}

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

tan (x) - 1 = 0, 0 \leq x \leq 2 π

1

を右側に移動します

tan (x) - 1 = 0

両辺に

1

を足す

tan (x) - 1 + 1 = 0 + 1

簡素化

tan (x) = 1

tan (x) = 1

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}

すべての解を組み合わせる

x = 0, x = π, x = 2 π, x = \frac{3 π}{4}, x = \frac{7 π}{4}, x = \frac{π}{4}, x = \frac{5 π}{4}

グラフ

人気の例

- 3 sin (t) = 0

3cos^2(θ)+7cos(θ)-8=0

3 cos^{2} (θ) + 7 cos (θ) - 8 = 0

2cos(2a)=sqrt(2),0<= θ<= 2pi

2 cos (2 a) = 2, 0 \leq θ \leq 2 π

solvefor t,y^4-81y=sin(t)

so l v e f or t, y^{4} - 81 y = sin (t)

11sin(B)-1=3sin(B)-1

11 sin (B) - 1 = 3 sin (B) - 1