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人気のある三角関数 >

tan(x)= 1/(sec(x))

解

tan (x) = \frac{1}{sec ( x )}

解

x = 0.66623 \dots + 2 πn, x = π - 0.66623 \dots + 2 πn

+1

度

x = 38.17270 \dots^{\circ} + 36 0^{\circ} n, x = 141.82729 \dots^{\circ} + 36 0^{\circ} n

解答ステップ

tan (x) = \frac{1}{sec ( x )}

両辺から

\frac{1}{sec ( x )}

を引く

tan (x) - \frac{1}{sec ( x )} = 0

簡素化

tan (x) - \frac{1}{sec ( x )} : \frac{tan ( x ) sec ( x ) - 1}{sec ( x )}

tan (x) - \frac{1}{sec ( x )}

元を分数に変換する:

tan (x) = \frac{tan ( x ) sec ( x )}{sec ( x )}

= \frac{tan ( x ) sec ( x )}{sec ( x )} - \frac{1}{sec ( 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{tan ( x ) sec ( x ) - 1}{sec ( x )}

\frac{tan ( x ) sec ( x ) - 1}{sec ( x )} = 0

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

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

サイン, コサインで表わす

- 1 + sec (x) tan (x)

基本的な三角関数の公式を使用する:

sec (x) = \frac{1}{cos ( x )}

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

基本的な三角関数の公式を使用する:

tan (x) = \frac{sin ( x )}{cos ( x )}

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

簡素化

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

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

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

\frac{1}{cos ( x )} \cdot \frac{sin ( x )}{cos ( x )}

分数を乗じる:

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

= \frac{1 \cdot sin ( x )}{cos ( x ) cos ( x )}

乗算:

1 \cdot sin (x) = sin (x)

= \frac{sin ( x )}{cos ( x ) cos ( x )}

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

cos (x) cos (x)

指数の規則を適用する:

a^{b} \cdot a^{c} = a^{b + c}

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

= cos^{1 + 1} (x)

数を足す:

1 + 1 = 2

= cos^{2} (x)

= \frac{sin ( x )}{cos ^{2} ( x )}

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

元を分数に変換する:

1 = \frac{1 cos ^{2} ( x )}{cos ^{2} ( x )}

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

乗算:

1 \cdot cos^{2} (x) = cos^{2} (x)

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

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

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

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

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

三角関数の公式を使用して書き換える

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

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

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

括弧を分配する

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

マイナス・プラスの規則を適用する

- (- a) = a, - (a) = - a

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

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

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

置換で解く

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

仮定:

sin (x) = u

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

- 1 + u + u^{2} = 0 : u = \frac{- 1 + 5}{2}, u = \frac{- 1 - 5}{2}

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

標準的な形式で書く

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

u^{2} + u - 1 = 0

解くとthe二次式

u^{2} + u - 1 = 0

二次Equationの公式:

次の場合:

a = 1, b = 1, c = - 1

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

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

1^{2} - 4 \cdot 1 \cdot (- 1) = 5

1^{2} - 4 \cdot 1 \cdot (- 1)

規則を適用

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 1 \cdot (- 1)

規則を適用

- (- a) = a

= 1 + 4 \cdot 1 \cdot 1

数を乗じる:

4 \cdot 1 \cdot 1 = 4

= 1 + 4

数を足す:

1 + 4 = 5

= 5

u_{1, 2} = \frac{- 1 \pm 5}{2 \cdot 1}

解を分離する

u_{1} = \frac{- 1 + 5}{2 \cdot 1}, u_{2} = \frac{- 1 - 5}{2 \cdot 1}

u = \frac{- 1 + 5}{2 \cdot 1} : \frac{- 1 + 5}{2}

\frac{- 1 + 5}{2 \cdot 1}

数を乗じる:

2 \cdot 1 = 2

= \frac{- 1 + 5}{2}

u = \frac{- 1 - 5}{2 \cdot 1} : \frac{- 1 - 5}{2}

\frac{- 1 - 5}{2 \cdot 1}

数を乗じる:

2 \cdot 1 = 2

= \frac{- 1 - 5}{2}

二次equationの解:

u = \frac{- 1 + 5}{2}, u = \frac{- 1 - 5}{2}

代用を戻す

u = sin (x)

sin (x) = \frac{- 1 + 5}{2}, sin (x) = \frac{- 1 - 5}{2}

sin (x) = \frac{- 1 + 5}{2}, sin (x) = \frac{- 1 - 5}{2}

sin (x) = \frac{- 1 + 5}{2} : x = arcsin (\frac{- 1 + 5}{2}) + 2 πn, x = π - arcsin (\frac{- 1 + 5}{2}) + 2 πn

sin (x) = \frac{- 1 + 5}{2}

三角関数の逆数プロパティを適用する

sin (x) = \frac{- 1 + 5}{2}

以下の一般解

sin (x) = \frac{- 1 + 5}{2}

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

x = arcsin (\frac{- 1 + 5}{2}) + 2 πn, x = π - arcsin (\frac{- 1 + 5}{2}) + 2 πn

x = arcsin (\frac{- 1 + 5}{2}) + 2 πn, x = π - arcsin (\frac{- 1 + 5}{2}) + 2 πn

sin (x) = \frac{- 1 - 5}{2} :

解なし

sin (x) = \frac{- 1 - 5}{2}

- 1 \leq sin (x) \leq 1

解なし

すべての解を組み合わせる

x = arcsin (\frac{- 1 + 5}{2}) + 2 πn, x = π - arcsin (\frac{- 1 + 5}{2}) + 2 πn

10進法形式で解を証明する

x = 0.66623 \dots + 2 πn, x = π - 0.66623 \dots + 2 πn

グラフ

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