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人気のある三角関数 >

12cos^2(x)+2cos(x)-2=0

解

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

解

x = 1.23095 \dots + 2 πn, x = 2 π - 1.23095 \dots + 2 πn, x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

+1

度

x = 70.52877 \dots^{\circ} + 36 0^{\circ} n, x = 289.47122 \dots^{\circ} + 36 0^{\circ} n, x = 12 0^{\circ} + 36 0^{\circ} n, x = 24 0^{\circ} + 36 0^{\circ} n

解答ステップ

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

置換で解く

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

仮定:

cos (x) = u

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

12 u^{2} + 2 u - 2 = 0 : u = \frac{1}{3}, u = - \frac{1}{2}

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

解くとthe二次式

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

二次Equationの公式:

次の場合:

a = 12, b = 2, c = - 2

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 12 ( - 2 )}{2 \cdot 12}

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 12 ( - 2 )}{2 \cdot 12}

2^{2} - 4 \cdot 12 (- 2) = 10

2^{2} - 4 \cdot 12 (- 2)

規則を適用

- (- a) = a

= 2^{2} + 4 \cdot 12 \cdot 2

数を乗じる:

4 \cdot 12 \cdot 2 = 96

= 2^{2} + 96

2^{2} = 4

= 4 + 96

数を足す:

4 + 96 = 100

= 100

数を因数に分解する:

100 = 1 0^{2}

= 1 0^{2}

累乗根の規則を適用する:

n a^{n} = a

1 0^{2} = 10

= 10

u_{1, 2} = \frac{- 2 \pm 10}{2 \cdot 12}

解を分離する

u_{1} = \frac{- 2 + 10}{2 \cdot 12}, u_{2} = \frac{- 2 - 10}{2 \cdot 12}

u = \frac{- 2 + 10}{2 \cdot 12} : \frac{1}{3}

\frac{- 2 + 10}{2 \cdot 12}

数を足す/引く:

- 2 + 10 = 8

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

数を乗じる:

2 \cdot 12 = 24

= \frac{8}{24}

共通因数を約分する:

8

= \frac{1}{3}

u = \frac{- 2 - 10}{2 \cdot 12} : - \frac{1}{2}

\frac{- 2 - 10}{2 \cdot 12}

数を引く:

- 2 - 10 = - 12

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

数を乗じる:

2 \cdot 12 = 24

= \frac{- 12}{24}

分数の規則を適用する:

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

= - \frac{12}{24}

共通因数を約分する:

12

= - \frac{1}{2}

二次equationの解:

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

代用を戻す

u = cos (x)

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

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

cos (x) = \frac{1}{3} : x = arccos (\frac{1}{3}) + 2 πn, x = 2 π - arccos (\frac{1}{3}) + 2 πn

cos (x) = \frac{1}{3}

三角関数の逆数プロパティを適用する

cos (x) = \frac{1}{3}

以下の一般解

cos (x) = \frac{1}{3}

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

x = arccos (\frac{1}{3}) + 2 πn, x = 2 π - arccos (\frac{1}{3}) + 2 πn

x = arccos (\frac{1}{3}) + 2 πn, x = 2 π - arccos (\frac{1}{3}) + 2 πn

cos (x) = - \frac{1}{2} : x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

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

以下の一般解

cos (x) = - \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}

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

すべての解を組み合わせる

x = arccos (\frac{1}{3}) + 2 πn, x = 2 π - arccos (\frac{1}{3}) + 2 πn, x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

10進法形式で解を証明する

x = 1.23095 \dots + 2 πn, x = 2 π - 1.23095 \dots + 2 πn, x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

グラフ

人気の例

sin (θ) = - 0.94

sin(θ)=0,0<= θ<= 2pi

sin (θ) = 0, 0 \leq θ \leq 2 π

sin (θ) = - 0.71

2 cos (4 x) = 0

tan(2θ)=(-1)/1

tan (2 θ) = \frac{- 1}{1}