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人気のある三角関数 >

4cos(2θ)-10cos(θ)+14=7,0<= θ<360

解

4 cos (2 θ) - 10 cos (θ) + 14 = 7, 0 \leq θ < 36 0^{\circ}

解

θ = 0.72273 \dots, θ = 36 0^{\circ} - 0.72273 \dots, θ = 6 0^{\circ}, θ = 30 0^{\circ}

+1

ラジアン

θ = 0.72273 \dots, θ = 2 π - 0.72273 \dots, θ = \frac{π}{3}, θ = \frac{5 π}{3}

解答ステップ

4 cos (2 θ) - 10 cos (θ) + 14 = 7, 0 \leq θ < 36 0^{\circ}

両辺から

7

を引く

4 cos (2 θ) - 10 cos (θ) + 7 = 0

三角関数の公式を使用して書き換える

7 - 10 cos (θ) + 4 cos (2 θ)

2倍角の公式を使用:

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

= 7 - 10 cos (θ) + 4 (2 cos^{2} (θ) - 1)

簡素化

7 - 10 cos (θ) + 4 (2 cos^{2} (θ) - 1) : 8 cos^{2} (θ) - 10 cos (θ) + 3

7 - 10 cos (θ) + 4 (2 cos^{2} (θ) - 1)

拡張

4 (2 cos^{2} (θ) - 1) : 8 cos^{2} (θ) - 4

4 (2 cos^{2} (θ) - 1)

分配法則を適用する:

a (b - c) = ab - a c

a = 4, b = 2 cos^{2} (θ), c = 1

= 4 \cdot 2 cos^{2} (θ) - 4 \cdot 1

簡素化

4 \cdot 2 cos^{2} (θ) - 4 \cdot 1 : 8 cos^{2} (θ) - 4

4 \cdot 2 cos^{2} (θ) - 4 \cdot 1

数を乗じる:

4 \cdot 2 = 8

= 8 cos^{2} (θ) - 4 \cdot 1

数を乗じる:

4 \cdot 1 = 4

= 8 cos^{2} (θ) - 4

= 8 cos^{2} (θ) - 4

= 7 - 10 cos (θ) + 8 cos^{2} (θ) - 4

簡素化

7 - 10 cos (θ) + 8 cos^{2} (θ) - 4 : 8 cos^{2} (θ) - 10 cos (θ) + 3

7 - 10 cos (θ) + 8 cos^{2} (θ) - 4

条件のようなグループ

= - 10 cos (θ) + 8 cos^{2} (θ) + 7 - 4

数を足す/引く:

7 - 4 = 3

= 8 cos^{2} (θ) - 10 cos (θ) + 3

= 8 cos^{2} (θ) - 10 cos (θ) + 3

= 8 cos^{2} (θ) - 10 cos (θ) + 3

3 - 10 cos (θ) + 8 cos^{2} (θ) = 0

置換で解く

3 - 10 cos (θ) + 8 cos^{2} (θ) = 0

仮定:

cos (θ) = u

3 - 10 u + 8 u^{2} = 0

3 - 10 u + 8 u^{2} = 0 : u = \frac{3}{4}, u = \frac{1}{2}

3 - 10 u + 8 u^{2} = 0

標準的な形式で書く

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

8 u^{2} - 10 u + 3 = 0

解くとthe二次式

8 u^{2} - 10 u + 3 = 0

二次Equationの公式:

次の場合:

a = 8, b = - 10, c = 3

u_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 8 \cdot 3}{2 \cdot 8}

u_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 8 \cdot 3}{2 \cdot 8}

(- 10)^{2} - 4 \cdot 8 \cdot 3 = 2

(- 10)^{2} - 4 \cdot 8 \cdot 3

指数の規則を適用する:

n

が偶数であれば

(- a)^{n} = a^{n}

(- 10)^{2} = 1 0^{2}

= 1 0^{2} - 4 \cdot 8 \cdot 3

数を乗じる:

4 \cdot 8 \cdot 3 = 96

= 1 0^{2} - 96

1 0^{2} = 100

= 100 - 96

数を引く:

100 - 96 = 4

= 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

u_{1, 2} = \frac{- ( - 10 ) \pm 2}{2 \cdot 8}

解を分離する

u_{1} = \frac{- ( - 10 ) + 2}{2 \cdot 8}, u_{2} = \frac{- ( - 10 ) - 2}{2 \cdot 8}

u = \frac{- ( - 10 ) + 2}{2 \cdot 8} : \frac{3}{4}

\frac{- ( - 10 ) + 2}{2 \cdot 8}

規則を適用

- (- a) = a

= \frac{10 + 2}{2 \cdot 8}

数を足す:

10 + 2 = 12

= \frac{12}{2 \cdot 8}

数を乗じる:

2 \cdot 8 = 16

= \frac{12}{16}

共通因数を約分する:

4

= \frac{3}{4}

u = \frac{- ( - 10 ) - 2}{2 \cdot 8} : \frac{1}{2}

\frac{- ( - 10 ) - 2}{2 \cdot 8}

規則を適用

- (- a) = a

= \frac{10 - 2}{2 \cdot 8}

数を引く:

10 - 2 = 8

= \frac{8}{2 \cdot 8}

数を乗じる:

2 \cdot 8 = 16

= \frac{8}{16}

共通因数を約分する:

8

= \frac{1}{2}

二次equationの解:

u = \frac{3}{4}, u = \frac{1}{2}

代用を戻す

u = cos (θ)

cos (θ) = \frac{3}{4}, cos (θ) = \frac{1}{2}

cos (θ) = \frac{3}{4}, cos (θ) = \frac{1}{2}

cos (θ) = \frac{3}{4}, 0 \leq θ < 36 0^{\circ} : θ = arccos (\frac{3}{4}), θ = 36 0^{\circ} - arccos (\frac{3}{4})

cos (θ) = \frac{3}{4}, 0 \leq θ < 36 0^{\circ}

三角関数の逆数プロパティを適用する

cos (θ) = \frac{3}{4}

以下の一般解

cos (θ) = \frac{3}{4}

cos (x) = a \Rightarrow x = arccos (a) + 36 0^{\circ} n, x = 36 0^{\circ} - arccos (a) + 36 0^{\circ} n

θ = arccos (\frac{3}{4}) + 36 0^{\circ} n, θ = 36 0^{\circ} - arccos (\frac{3}{4}) + 36 0^{\circ} n

θ = arccos (\frac{3}{4}) + 36 0^{\circ} n, θ = 36 0^{\circ} - arccos (\frac{3}{4}) + 36 0^{\circ} n

範囲の解答

0 \leq θ < 36 0^{\circ}

θ = arccos (\frac{3}{4}), θ = 36 0^{\circ} - arccos (\frac{3}{4})

cos (θ) = \frac{1}{2}, 0 \leq θ < 36 0^{\circ} : θ = 6 0^{\circ}, θ = 30 0^{\circ}

cos (θ) = \frac{1}{2}, 0 \leq θ < 36 0^{\circ}

以下の一般解

cos (θ) = \frac{1}{2}

cos (x)

36 0^{\circ} n

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

θ = 6 0^{\circ} + 36 0^{\circ} n, θ = 30 0^{\circ} + 36 0^{\circ} n

θ = 6 0^{\circ} + 36 0^{\circ} n, θ = 30 0^{\circ} + 36 0^{\circ} n

範囲の解答

0 \leq θ < 36 0^{\circ}

θ = 6 0^{\circ}, θ = 30 0^{\circ}

すべての解を組み合わせる

θ = arccos (\frac{3}{4}), θ = 36 0^{\circ} - arccos (\frac{3}{4}), θ = 6 0^{\circ}, θ = 30 0^{\circ}

10進法形式で解を証明する

θ = 0.72273 \dots, θ = 36 0^{\circ} - 0.72273 \dots, θ = 6 0^{\circ}, θ = 30 0^{\circ}

グラフ

人気の例

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