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人気のある三角関数 >

3cos(2x)-2cos^2(x)-5=2cos^2(x)-7

解

3 cos (2 x) - 2 cos^{2} (x) - 5 = 2 cos^{2} (x) - 7

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

3 cos (2 x) - 2 cos^{2} (x) - 5 = 2 cos^{2} (x) - 7

両辺から

2 cos^{2} (x) - 7

を引く

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

三角関数の公式を使用して書き換える

2 + 3 cos (2 x) - 4 cos^{2} (x)

2倍角の公式を使用:

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

= 2 + 3 (2 cos^{2} (x) - 1) - 4 cos^{2} (x)

簡素化

2 + 3 (2 cos^{2} (x) - 1) - 4 cos^{2} (x) : 2 cos^{2} (x) - 1

2 + 3 (2 cos^{2} (x) - 1) - 4 cos^{2} (x)

拡張

3 (2 cos^{2} (x) - 1) : 6 cos^{2} (x) - 3

3 (2 cos^{2} (x) - 1)

分配法則を適用する:

a (b - c) = ab - a c

a = 3, b = 2 cos^{2} (x), c = 1

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

簡素化

3 \cdot 2 cos^{2} (x) - 3 \cdot 1 : 6 cos^{2} (x) - 3

3 \cdot 2 cos^{2} (x) - 3 \cdot 1

数を乗じる:

3 \cdot 2 = 6

= 6 cos^{2} (x) - 3 \cdot 1

数を乗じる:

3 \cdot 1 = 3

= 6 cos^{2} (x) - 3

= 6 cos^{2} (x) - 3

= 2 + 6 cos^{2} (x) - 3 - 4 cos^{2} (x)

簡素化

2 + 6 cos^{2} (x) - 3 - 4 cos^{2} (x) : 2 cos^{2} (x) - 1

2 + 6 cos^{2} (x) - 3 - 4 cos^{2} (x)

条件のようなグループ

= 6 cos^{2} (x) - 4 cos^{2} (x) + 2 - 3

類似した元を足す:

6 cos^{2} (x) - 4 cos^{2} (x) = 2 cos^{2} (x)

= 2 cos^{2} (x) + 2 - 3

数を足す/引く:

2 - 3 = - 1

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

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

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

- 1 + 2 cos^{2} (x) = 0

置換で解く

- 1 + 2 cos^{2} (x) = 0

仮定:

cos (x) = u

- 1 + 2 u^{2} = 0

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

- 1 + 2 u^{2} = 0

1

を右側に移動します

- 1 + 2 u^{2} = 0

両辺に

1

を足す

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

簡素化

2 u^{2} = 1

2 u^{2} = 1

以下で両辺を割る

2

2 u^{2} = 1

以下で両辺を割る

2

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

簡素化

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

u^{2} = \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}

代用を戻す

u = cos (x)

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

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

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

sin(3x+48)=cos(10-x)

sin (3 x + 48) = cos (1 0^{\circ} - x)

2sin(θ/3)+sqrt(3)=0,0<= θ<= 2pi

2 sin (\frac{θ}{3}) + 3 = 0, 0 \leq θ \leq 2 π

sin(2x)-sqrt(2)sin(x)=0,0<= x<= 2pi

sin (2 x) - 2 sin (x) = 0, 0 \leq x \leq 2 π

2sin(x-pi/3)+sqrt(3)=0

2 sin (x - \frac{π}{3}) + 3 = 0

sin (x) = 0.77