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人気のある三角関数 >

tan(2x)+sec(2x)=11

解

tan (2 x) + sec (2 x) = 11

解

x = \frac{1.38947 \dots}{2} + πn

+1

度

x = 39.80557 \dots^{\circ} + 18 0^{\circ} n

解答ステップ

tan (2 x) + sec (2 x) = 11

両辺から

11

を引く

tan (2 x) + sec (2 x) - 11 = 0

サイン, コサインで表わす

\frac{sin ( 2 x )}{cos ( 2 x )} + \frac{1}{cos ( 2 x )} - 11 = 0

簡素化

\frac{sin ( 2 x )}{cos ( 2 x )} + \frac{1}{cos ( 2 x )} - 11 : \frac{sin ( 2 x ) + 1 - 11 cos ( 2 x )}{cos ( 2 x )}

\frac{sin ( 2 x )}{cos ( 2 x )} + \frac{1}{cos ( 2 x )} - 11

分数を組み合わせる

\frac{sin ( 2 x )}{cos ( 2 x )} + \frac{1}{cos ( 2 x )} : \frac{sin ( 2 x ) + 1}{cos ( 2 x )}

規則を適用

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{sin ( 2 x ) + 1}{cos ( 2 x )}

= \frac{sin ( 2 x ) + 1}{cos ( 2 x )} - 11

元を分数に変換する:

11 = \frac{11 cos ( 2 x )}{cos ( 2 x )}

= \frac{sin ( 2 x ) + 1}{cos ( 2 x )} - \frac{11 cos ( 2 x )}{cos ( 2 x )}

分母が等しいので, 分数を組み合わせる:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{sin ( 2 x ) + 1 - 11 cos ( 2 x )}{cos ( 2 x )}

\frac{sin ( 2 x ) + 1 - 11 cos ( 2 x )}{cos ( 2 x )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

sin (2 x) + 1 - 11 cos (2 x) = 0

両辺に

11 cos (2 x)

を足す

sin (2 x) + 1 = 11 cos (2 x)

両辺を2乗する

(sin (2 x) + 1)^{2} = (11 cos (2 x))^{2}

両辺から

(11 cos (2 x))^{2}

を引く

(sin (2 x) + 1)^{2} - 121 cos^{2} (2 x) = 0

三角関数の公式を使用して書き換える

(1 + sin (2 x))^{2} - 121 cos^{2} (2 x)

ピタゴラスの公式を使用する:

cos^{2} (x) + sin^{2} (x) = 1

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

= (1 + sin (2 x))^{2} - 121 (1 - sin^{2} (2 x))

簡素化

(1 + sin (2 x))^{2} - 121 (1 - sin^{2} (2 x)) : 122 sin^{2} (2 x) + 2 sin (2 x) - 120

(1 + sin (2 x))^{2} - 121 (1 - sin^{2} (2 x))

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

完全平方式を適用する:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

a = 1, b = sin (2 x)

= 1^{2} + 2 \cdot 1 \cdot sin (2 x) + sin^{2} (2 x)

簡素化

1^{2} + 2 \cdot 1 \cdot sin (2 x) + sin^{2} (2 x) : 1 + 2 sin (2 x) + sin^{2} (2 x)

1^{2} + 2 \cdot 1 \cdot sin (2 x) + sin^{2} (2 x)

規則を適用

1^{a} = 1

1^{2} = 1

= 1 + 2 \cdot 1 \cdot sin (2 x) + sin^{2} (2 x)

数を乗じる:

2 \cdot 1 = 2

= 1 + 2 sin (2 x) + sin^{2} (2 x)

= 1 + 2 sin (2 x) + sin^{2} (2 x)

= 1 + 2 sin (2 x) + sin^{2} (2 x) - 121 (1 - sin^{2} (2 x))

拡張

- 121 (1 - sin^{2} (2 x)) : - 121 + 121 sin^{2} (2 x)

- 121 (1 - sin^{2} (2 x))

分配法則を適用する:

a (b - c) = ab - a c

a = - 121, b = 1, c = sin^{2} (2 x)

= - 121 \cdot 1 - (- 121) sin^{2} (2 x)

マイナス・プラスの規則を適用する

- (- a) = a

= - 121 \cdot 1 + 121 sin^{2} (2 x)

数を乗じる:

121 \cdot 1 = 121

= - 121 + 121 sin^{2} (2 x)

= 1 + 2 sin (2 x) + sin^{2} (2 x) - 121 + 121 sin^{2} (2 x)

簡素化

1 + 2 sin (2 x) + sin^{2} (2 x) - 121 + 121 sin^{2} (2 x) : 122 sin^{2} (2 x) + 2 sin (2 x) - 120

1 + 2 sin (2 x) + sin^{2} (2 x) - 121 + 121 sin^{2} (2 x)

条件のようなグループ

= 2 sin (2 x) + sin^{2} (2 x) + 121 sin^{2} (2 x) + 1 - 121

類似した元を足す:

sin^{2} (2 x) + 121 sin^{2} (2 x) = 122 sin^{2} (2 x)

= 2 sin (2 x) + 122 sin^{2} (2 x) + 1 - 121

数を足す/引く:

1 - 121 = - 120

= 122 sin^{2} (2 x) + 2 sin (2 x) - 120

= 122 sin^{2} (2 x) + 2 sin (2 x) - 120

= 122 sin^{2} (2 x) + 2 sin (2 x) - 120

- 120 + 122 sin^{2} (2 x) + 2 sin (2 x) = 0

置換で解く

- 120 + 122 sin^{2} (2 x) + 2 sin (2 x) = 0

仮定:

sin (2 x) = u

- 120 + 122 u^{2} + 2 u = 0

- 120 + 122 u^{2} + 2 u = 0 : u = \frac{60}{61}, u = - 1

- 120 + 122 u^{2} + 2 u = 0

標準的な形式で書く

a x^{2} + b x + c = 0

122 u^{2} + 2 u - 120 = 0

解くとthe二次式

122 u^{2} + 2 u - 120 = 0

二次Equationの公式:

次の場合:

a = 122, b = 2, c = - 120

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 122 ( - 120 )}{2 \cdot 122}

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 122 ( - 120 )}{2 \cdot 122}

2^{2} - 4 \cdot 122 (- 120) = 242

2^{2} - 4 \cdot 122 (- 120)

規則を適用

- (- a) = a

= 2^{2} + 4 \cdot 122 \cdot 120

数を乗じる:

4 \cdot 122 \cdot 120 = 58560

= 2^{2} + 58560

2^{2} = 4

= 4 + 58560

数を足す:

4 + 58560 = 58564

= 58564

数を因数に分解する:

58564 = 24 2^{2}

= 24 2^{2}

累乗根の規則を適用する:

n a^{n} = a

24 2^{2} = 242

= 242

u_{1, 2} = \frac{- 2 \pm 242}{2 \cdot 122}

解を分離する

u_{1} = \frac{- 2 + 242}{2 \cdot 122}, u_{2} = \frac{- 2 - 242}{2 \cdot 122}

u = \frac{- 2 + 242}{2 \cdot 122} : \frac{60}{61}

\frac{- 2 + 242}{2 \cdot 122}

数を足す/引く:

- 2 + 242 = 240

= \frac{240}{2 \cdot 122}

数を乗じる:

2 \cdot 122 = 244

= \frac{240}{244}

共通因数を約分する:

4

= \frac{60}{61}

u = \frac{- 2 - 242}{2 \cdot 122} : - 1

\frac{- 2 - 242}{2 \cdot 122}

数を引く:

- 2 - 242 = - 244

= \frac{- 244}{2 \cdot 122}

数を乗じる:

2 \cdot 122 = 244

= \frac{- 244}{244}

分数の規則を適用する:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{244}{244}

規則を適用

\frac{a}{a} = 1

= - 1

二次equationの解:

u = \frac{60}{61}, u = - 1

代用を戻す

u = sin (2 x)

sin (2 x) = \frac{60}{61}, sin (2 x) = - 1

sin (2 x) = \frac{60}{61}, sin (2 x) = - 1

sin (2 x) = \frac{60}{61} : x = \frac{arcsin ( \frac{60}{61} )}{2} + πn, x = \frac{π}{2} - \frac{arcsin ( \frac{60}{61} )}{2} + πn

sin (2 x) = \frac{60}{61}

三角関数の逆数プロパティを適用する

sin (2 x) = \frac{60}{61}

以下の一般解

sin (2 x) = \frac{60}{61}

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

2 x = arcsin (\frac{60}{61}) + 2 πn, 2 x = π - arcsin (\frac{60}{61}) + 2 πn

2 x = arcsin (\frac{60}{61}) + 2 πn, 2 x = π - arcsin (\frac{60}{61}) + 2 πn

解く

2 x = arcsin (\frac{60}{61}) + 2 πn : x = \frac{arcsin ( \frac{60}{61} )}{2} + πn

2 x = arcsin (\frac{60}{61}) + 2 πn

以下で両辺を割る

2

2 x = arcsin (\frac{60}{61}) + 2 πn

以下で両辺を割る

2

\frac{2 x}{2} = \frac{arcsin ( \frac{60}{61} )}{2} + \frac{2 πn}{2}

簡素化

x = \frac{arcsin ( \frac{60}{61} )}{2} + πn

x = \frac{arcsin ( \frac{60}{61} )}{2} + πn

解く

2 x = π - arcsin (\frac{60}{61}) + 2 πn : x = \frac{π}{2} - \frac{arcsin ( \frac{60}{61} )}{2} + πn

2 x = π - arcsin (\frac{60}{61}) + 2 πn

以下で両辺を割る

2

2 x = π - arcsin (\frac{60}{61}) + 2 πn

以下で両辺を割る

2

\frac{2 x}{2} = \frac{π}{2} - \frac{arcsin ( \frac{60}{61} )}{2} + \frac{2 πn}{2}

簡素化

x = \frac{π}{2} - \frac{arcsin ( \frac{60}{61} )}{2} + πn

x = \frac{π}{2} - \frac{arcsin ( \frac{60}{61} )}{2} + πn

x = \frac{arcsin ( \frac{60}{61} )}{2} + πn, x = \frac{π}{2} - \frac{arcsin ( \frac{60}{61} )}{2} + πn

sin (2 x) = - 1 : x = \frac{3 π}{4} + πn

sin (2 x) = - 1

以下の一般解

sin (2 x) = - 1

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}

2 x = \frac{3 π}{2} + 2 πn

2 x = \frac{3 π}{2} + 2 πn

解く

2 x = \frac{3 π}{2} + 2 πn : x = \frac{3 π}{4} + πn

2 x = \frac{3 π}{2} + 2 πn

以下で両辺を割る

2

2 x = \frac{3 π}{2} + 2 πn

以下で両辺を割る

2

\frac{2 x}{2} = \frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2}

簡素化

\frac{2 x}{2} = \frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2}

簡素化

\frac{2 x}{2} : x

\frac{2 x}{2}

数を割る:

\frac{2}{2} = 1

= x

簡素化

\frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2} : \frac{3 π}{4} + πn

\frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2}

\frac{\frac{3 π}{2}}{2} = \frac{3 π}{4}

\frac{\frac{3 π}{2}}{2}

分数の規則を適用する:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{3 π}{2 \cdot 2}

数を乗じる:

2 \cdot 2 = 4

= \frac{3 π}{4}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数を割る:

\frac{2}{2} = 1

= πn

= \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

すべての解を組み合わせる

x = \frac{arcsin ( \frac{60}{61} )}{2} + πn, x = \frac{π}{2} - \frac{arcsin ( \frac{60}{61} )}{2} + πn, x = \frac{3 π}{4} + πn

元のequationに当てはめて解を検算する

tan (2 x) + sec (2 x) = 11

に当てはめて解を確認する

equationに一致しないものを削除する。

解答を確認する

\frac{arcsin ( \frac{60}{61} )}{2} + πn :

真

\frac{arcsin ( \frac{60}{61} )}{2} + πn

挿入

n = 1

\frac{arcsin ( \frac{60}{61} )}{2} + π 1

tan (2 x) + sec (2 x) = 11

の挿入向け

x = \frac{arcsin ( \frac{60}{61} )}{2} + π 1

tan (2 (\frac{arcsin ( \frac{60}{61} )}{2} + π 1)) + sec (2 (\frac{arcsin ( \frac{60}{61} )}{2} + π 1)) = 11

改良

11 = 11

\Rightarrow 真

解答を確認する

\frac{π}{2} - \frac{arcsin ( \frac{60}{61} )}{2} + πn :

偽

\frac{π}{2} - \frac{arcsin ( \frac{60}{61} )}{2} + πn

挿入

n = 1

\frac{π}{2} - \frac{arcsin ( \frac{60}{61} )}{2} + π 1

tan (2 x) + sec (2 x) = 11

の挿入向け

x = \frac{π}{2} - \frac{arcsin ( \frac{60}{61} )}{2} + π 1

tan (2 (\frac{π}{2} - \frac{arcsin ( \frac{60}{61} )}{2} + π 1)) + sec (2 (\frac{π}{2} - \frac{arcsin ( \frac{60}{61} )}{2} + π 1)) = 11

改良

- 11 = 11

\Rightarrow 偽

解答を確認する

\frac{3 π}{4} + πn :

偽

\frac{3 π}{4} + πn

挿入

n = 1

\frac{3 π}{4} + π 1

tan (2 x) + sec (2 x) = 11

の挿入向け

x = \frac{3 π}{4} + π 1

tan (2 (\frac{3 π}{4} + π 1)) + sec (2 (\frac{3 π}{4} + π 1)) = 11

未定義

\Rightarrow 偽

x = \frac{arcsin ( \frac{60}{61} )}{2} + πn

10進法形式で解を証明する

x = \frac{1.38947 \dots}{2} + πn

グラフ

人気の例

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sin(θ)=1+cos(θ)

sin (θ) = 1 + cos (θ)

7cos(2x)=7cos^2(x)-3

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