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人気のある三角関数 >

4cos^4(x)-1=0

解

4 cos^{4} (x) - 1 = 0

解

x = 0.78539 \dots + 2 πn, x = 2 π - 0.78539 \dots + 2 πn, x = 2.35619 \dots + 2 πn, x = - 2.35619 \dots + 2 πn

+1

度

x = 4 5^{\circ} + 36 0^{\circ} n, x = 31 5^{\circ} + 36 0^{\circ} n, x = 13 5^{\circ} + 36 0^{\circ} n, x = - 13 5^{\circ} + 36 0^{\circ} n

解答ステップ

4 cos^{4} (x) - 1 = 0

置換で解く

4 cos^{4} (x) - 1 = 0

仮定:

cos (x) = u

4 u^{4} - 1 = 0

4 u^{4} - 1 = 0 : u = \frac{1}{2}, u = - \frac{1}{2}, u = i \frac{1}{2}, u = - i \frac{1}{2}

4 u^{4} - 1 = 0

1

を右側に移動します

4 u^{4} - 1 = 0

両辺に

1

を足す

4 u^{4} - 1 + 1 = 0 + 1

簡素化

4 u^{4} = 1

4 u^{4} = 1

以下で両辺を割る

4

4 u^{4} = 1

以下で両辺を割る

4

\frac{4 u ^{4}}{4} = \frac{1}{4}

簡素化

u^{4} = \frac{1}{4}

u^{4} = \frac{1}{4}

equationを

v = u^{2}

と以下で書き換える:

v^{2} = u^{4}

v^{2} = \frac{1}{4}

解く

v^{2} = \frac{1}{4} : v = \frac{1}{4}, v = - \frac{1}{4}

v^{2} = \frac{1}{4}

(g (x))^{2} = f (a)

の場合, 解は

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

v = \frac{1}{4}, v = - \frac{1}{4}

v = \frac{1}{4}, v = - \frac{1}{4}

再び

v = u^{2}

に置き換えて以下を解く:

u

解く

u^{2} = \frac{1}{4} : u = \frac{1}{2}, u = - \frac{1}{2}

u^{2} = \frac{1}{4}

簡素化

\frac{1}{4} : \frac{1}{2}

\frac{1}{4}

累乗根の規則を適用する:

n \frac{a}{b} = \frac{n a}{n b},

, 以下を想定

a \geq 0, b \geq 0

= \frac{1}{4}

4 = 2

4

数を因数に分解する:

4 = 2^{2}

= 2^{2}

累乗根の規則を適用する:

n a^{n} = a

2^{2} = 2

= 2

= \frac{1}{2}

規則を適用

1 = 1

= \frac{1}{2}

x^{2} = f (a)

の場合, 解は

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

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

解く

u^{2} = - \frac{1}{4} : u = i \frac{1}{2}, u = - i \frac{1}{2}

u^{2} = - \frac{1}{4}

簡素化

- \frac{1}{4} : - \frac{1}{2}

- \frac{1}{4}

簡素化

\frac{1}{4} : \frac{1}{2}

\frac{1}{4}

累乗根の規則を適用する:

n \frac{a}{b} = \frac{n a}{n b},

, 以下を想定

a \geq 0, b \geq 0

= \frac{1}{4}

4 = 2

4

数を因数に分解する:

4 = 2^{2}

= 2^{2}

累乗根の規則を適用する:

n a^{n} = a

2^{2} = 2

= 2

= \frac{1}{2}

= - \frac{1}{2}

規則を適用

1 = 1

= - \frac{1}{2}

x^{2} = f (a)

の場合, 解は

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

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

簡素化

- \frac{1}{2} : i \frac{1}{2}

- \frac{1}{2}

累乗根の規則を適用する:

- a = - 1 a

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

= - 1 \frac{1}{2}

虚数の規則を適用する:

- 1 = i

= i \frac{1}{2}

簡素化

- - \frac{1}{2} : - i \frac{1}{2}

- - \frac{1}{2}

簡素化

- \frac{1}{2} : i \frac{1}{2}

- \frac{1}{2}

累乗根の規則を適用する:

- a = - 1 a

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

= - 1 \frac{1}{2}

虚数の規則を適用する:

- 1 = i

= i \frac{1}{2}

= - i \frac{1}{2}

u = i \frac{1}{2}, u = - i \frac{1}{2}

解答は

u = \frac{1}{2}, u = - \frac{1}{2}, u = i \frac{1}{2}, u = - i \frac{1}{2}

代用を戻す

u = cos (x)

cos (x) = \frac{1}{2}, cos (x) = - \frac{1}{2}, cos (x) = i \frac{1}{2}, cos (x) = - i \frac{1}{2}

cos (x) = \frac{1}{2}, cos (x) = - \frac{1}{2}, cos (x) = i \frac{1}{2}, cos (x) = - i \frac{1}{2}

cos (x) = \frac{1}{2} : x = arccos (\frac{1}{2}) + 2 πn, x = 2 π - arccos (\frac{1}{2}) + 2 πn

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

三角関数の逆数プロパティを適用する

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

以下の一般解

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

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

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

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

cos (x) = - \frac{1}{2} : x = arccos (- \frac{1}{2}) + 2 πn, x = - arccos (- \frac{1}{2}) + 2 πn

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

三角関数の逆数プロパティを適用する

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

以下の一般解

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

cos (x) = - a \Rightarrow x = arccos (- a) + 2 πn, x = - arccos (- a) + 2 πn

x = arccos (- \frac{1}{2}) + 2 πn, x = - arccos (- \frac{1}{2}) + 2 πn

x = arccos (- \frac{1}{2}) + 2 πn, x = - arccos (- \frac{1}{2}) + 2 πn

cos (x) = i \frac{1}{2} :

解なし

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

解なし

cos (x) = - i \frac{1}{2} :

解なし

cos (x) = - i \frac{1}{2}

解なし

すべての解を組み合わせる

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

10進法形式で解を証明する

x = 0.78539 \dots + 2 πn, x = 2 π - 0.78539 \dots + 2 πn, x = 2.35619 \dots + 2 πn, x = - 2.35619 \dots + 2 πn

グラフ

人気の例

arccos(2x)-arccos(x)= pi/3

arccos (2 x) - arccos (x) = \frac{π}{3}

cos(2θ)= 1/2 ,0<= θ<= 2pi

cos (2 θ) = \frac{1}{2}, 0 \leq θ \leq 2 π

tan(x-pi/2)=(sqrt(3))/3

tan (x - \frac{π}{2}) = \frac{3}{3}

2 = 2 cos (3 x)

cos(2x)=sqrt(3)

cos (2 x) = 3