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人気のある >

2cos^2(θ)+cos(θ)-1=0,\forall 0<= θ<2pi

解

2 cos^{2} (θ) + cos (θ) - 1 = 0, \forall 0 \leq θ < 2 π

解

θ = \frac{π}{3}, θ = \frac{5 π}{3}, θ = π

+1

度

θ = 6 0^{\circ}, θ = 30 0^{\circ}, θ = 18 0^{\circ}

解答ステップ

2 cos^{2} (θ) + cos (θ) - 1 = 0, \forall 0 \leq θ < 2 π

置換で解く

2 cos^{2} (θ) + cos (θ) - 1 = 0

仮定:

cos (θ) = u

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

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

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

解くとthe二次式

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

二次Equationの公式:

次の場合:

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

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

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

1^{2} - 4 \cdot 2 (- 1) = 3

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

規則を適用

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 2 (- 1)

規則を適用

- (- a) = a

= 1 + 4 \cdot 2 \cdot 1

数を乗じる:

4 \cdot 2 \cdot 1 = 8

= 1 + 8

数を足す:

1 + 8 = 9

= 9

数を因数に分解する:

9 = 3^{2}

= 3^{2}

累乗根の規則を適用する:

3^{2} = 3

= 3

u_{1, 2} = \frac{- 1 \pm 3}{2 \cdot 2}

解を分離する

u_{1} = \frac{- 1 + 3}{2 \cdot 2}, u_{2} = \frac{- 1 - 3}{2 \cdot 2}

u = \frac{- 1 + 3}{2 \cdot 2} : \frac{1}{2}

\frac{- 1 + 3}{2 \cdot 2}

数を足す/引く:

- 1 + 3 = 2

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

数を乗じる:

2 \cdot 2 = 4

= \frac{2}{4}

共通因数を約分する:

2

= \frac{1}{2}

u = \frac{- 1 - 3}{2 \cdot 2} : - 1

\frac{- 1 - 3}{2 \cdot 2}

数を引く:

- 1 - 3 = - 4

= \frac{- 4}{2 \cdot 2}

数を乗じる:

2 \cdot 2 = 4

= \frac{- 4}{4}

分数の規則を適用する:

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

= - \frac{4}{4}

規則を適用

\frac{a}{a} = 1

= - 1

二次equationの解:

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

代用を戻す

u = cos (θ)

cos (θ) = \frac{1}{2}, cos (θ) = - 1

cos (θ) = \frac{1}{2}, cos (θ) = - 1

cos (θ) = \frac{1}{2}, 0 \leq θ < 2 π : θ = \frac{π}{3}, θ = \frac{5 π}{3}

cos (θ) = \frac{1}{2}, 0 \leq θ < 2 π

以下の一般解

cos (θ) = \frac{1}{2}

cos (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} 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 = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

範囲の解答

0 \leq θ < 2 π

θ = \frac{π}{3}, θ = \frac{5 π}{3}

cos (θ) = - 1, 0 \leq θ < 2 π : θ = π

cos (θ) = - 1, 0 \leq θ < 2 π

以下の一般解

cos (θ) = - 1

cos (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} 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 = π + 2 πn

θ = π + 2 πn

範囲の解答

0 \leq θ < 2 π

θ = π

すべての解を組み合わせる

θ = \frac{π}{3}, θ = \frac{5 π}{3}, θ = π

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sin (θ) = 0.35

sin (θ) = 0.07

9sin^2(x)tan(x)=16tan(x)

9 sin^{2} (x) tan (x) = 16 tan (x)

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