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人気のある三角関数 >

4cos^2(x)+cos(x)-2=0

解

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

解

x = 0.93592 \dots + 2 πn, x = 2 π - 0.93592 \dots + 2 πn, x = 2.57376 \dots + 2 πn, x = - 2.57376 \dots + 2 πn

+1

度

x = 53.62480 \dots^{\circ} + 36 0^{\circ} n, x = 306.37519 \dots^{\circ} + 36 0^{\circ} n, x = 147.46577 \dots^{\circ} + 36 0^{\circ} n, x = - 147.46577 \dots^{\circ} + 36 0^{\circ} n

解答ステップ

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

置換で解く

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

仮定:

cos (x) = u

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

4 u^{2} + u - 2 = 0 : u = \frac{- 1 + 33}{8}, u = \frac{- 1 - 33}{8}

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

解くとthe二次式

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

二次Equationの公式:

次の場合:

a = 4, b = 1, c = - 2

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

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

1^{2} - 4 \cdot 4 (- 2) = 33

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

規則を適用

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 4 (- 2)

規則を適用

- (- a) = a

= 1 + 4 \cdot 4 \cdot 2

数を乗じる:

4 \cdot 4 \cdot 2 = 32

= 1 + 32

数を足す:

1 + 32 = 33

= 33

u_{1, 2} = \frac{- 1 \pm 33}{2 \cdot 4}

解を分離する

u_{1} = \frac{- 1 + 33}{2 \cdot 4}, u_{2} = \frac{- 1 - 33}{2 \cdot 4}

u = \frac{- 1 + 33}{2 \cdot 4} : \frac{- 1 + 33}{8}

\frac{- 1 + 33}{2 \cdot 4}

数を乗じる:

2 \cdot 4 = 8

= \frac{- 1 + 33}{8}

u = \frac{- 1 - 33}{2 \cdot 4} : \frac{- 1 - 33}{8}

\frac{- 1 - 33}{2 \cdot 4}

数を乗じる:

2 \cdot 4 = 8

= \frac{- 1 - 33}{8}

二次equationの解:

u = \frac{- 1 + 33}{8}, u = \frac{- 1 - 33}{8}

代用を戻す

u = cos (x)

cos (x) = \frac{- 1 + 33}{8}, cos (x) = \frac{- 1 - 33}{8}

cos (x) = \frac{- 1 + 33}{8}, cos (x) = \frac{- 1 - 33}{8}

cos (x) = \frac{- 1 + 33}{8} : x = arccos (\frac{- 1 + 33}{8}) + 2 πn, x = 2 π - arccos (\frac{- 1 + 33}{8}) + 2 πn

cos (x) = \frac{- 1 + 33}{8}

三角関数の逆数プロパティを適用する

cos (x) = \frac{- 1 + 33}{8}

以下の一般解

cos (x) = \frac{- 1 + 33}{8}

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

x = arccos (\frac{- 1 + 33}{8}) + 2 πn, x = 2 π - arccos (\frac{- 1 + 33}{8}) + 2 πn

x = arccos (\frac{- 1 + 33}{8}) + 2 πn, x = 2 π - arccos (\frac{- 1 + 33}{8}) + 2 πn

cos (x) = \frac{- 1 - 33}{8} : x = arccos (\frac{- 1 - 33}{8}) + 2 πn, x = - arccos (\frac{- 1 - 33}{8}) + 2 πn

cos (x) = \frac{- 1 - 33}{8}

三角関数の逆数プロパティを適用する

cos (x) = \frac{- 1 - 33}{8}

以下の一般解

cos (x) = \frac{- 1 - 33}{8}

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

x = arccos (\frac{- 1 - 33}{8}) + 2 πn, x = - arccos (\frac{- 1 - 33}{8}) + 2 πn

x = arccos (\frac{- 1 - 33}{8}) + 2 πn, x = - arccos (\frac{- 1 - 33}{8}) + 2 πn

すべての解を組み合わせる

x = arccos (\frac{- 1 + 33}{8}) + 2 πn, x = 2 π - arccos (\frac{- 1 + 33}{8}) + 2 πn, x = arccos (\frac{- 1 - 33}{8}) + 2 πn, x = - arccos (\frac{- 1 - 33}{8}) + 2 πn

10進法形式で解を証明する

x = 0.93592 \dots + 2 πn, x = 2 π - 0.93592 \dots + 2 πn, x = 2.57376 \dots + 2 πn, x = - 2.57376 \dots + 2 πn

グラフ

人気の例

tan(x)=-1+sqrt(2)

tan (x) = - 1 + 2

cot (θ) = - \frac{3}{4}

solvefor y,7e^x-sin(y)=x

so l v e f or y, 7 e^{x} - sin (y) = x

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sin (x) = \frac{10}{12}