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人気のある三角関数 >

-4cos^2(θ)=-6cos^2(θ)+1

解

- 4 cos^{2} (θ) = - 6 cos^{2} (θ) + 1

解

θ = 0.78539 \dots + 2 πn, θ = 2 π - 0.78539 \dots + 2 πn, θ = 2.35619 \dots + 2 πn, θ = - 2.35619 \dots + 2 πn

+1

度

θ = 4 5^{\circ} + 36 0^{\circ} n, θ = 31 5^{\circ} + 36 0^{\circ} n, θ = 13 5^{\circ} + 36 0^{\circ} n, θ = - 13 5^{\circ} + 36 0^{\circ} n

解答ステップ

- 4 cos^{2} (θ) = - 6 cos^{2} (θ) + 1

置換で解く

- 4 cos^{2} (θ) = - 6 cos^{2} (θ) + 1

仮定:

cos (θ) = u

- 4 u^{2} = - 6 u^{2} + 1

- 4 u^{2} = - 6 u^{2} + 1 : u = \frac{1}{2}, u = - \frac{1}{2}

- 4 u^{2} = - 6 u^{2} + 1

6 u^{2}

を左側に移動します

- 4 u^{2} = - 6 u^{2} + 1

両辺に

6 u^{2}

を足す

- 4 u^{2} + 6 u^{2} = - 6 u^{2} + 1 + 6 u^{2}

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 (θ)

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

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

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

cos (θ) = \frac{1}{2}

三角関数の逆数プロパティを適用する

cos (θ) = \frac{1}{2}

以下の一般解

cos (θ) = \frac{1}{2}

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

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

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

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

cos (θ) = - \frac{1}{2}

三角関数の逆数プロパティを適用する

cos (θ) = - \frac{1}{2}

以下の一般解

cos (θ) = - \frac{1}{2}

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

θ = arccos (- \frac{1}{2}) + 2 πn, θ = - arccos (- \frac{1}{2}) + 2 πn

θ = arccos (- \frac{1}{2}) + 2 πn, θ = - arccos (- \frac{1}{2}) + 2 πn

すべての解を組み合わせる

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

10進法形式で解を証明する

θ = 0.78539 \dots + 2 πn, θ = 2 π - 0.78539 \dots + 2 πn, θ = 2.35619 \dots + 2 πn, θ = - 2.35619 \dots + 2 πn

グラフ

人気の例

cos(4x)-3cos(2x)-1=0

cos (4 x) - 3 cos (2 x) - 1 = 0

7 sin (x) + 2 = 0

sin(x+30)=cos(x+60)

sin (x + 3 0^{\circ}) = cos (x + 6 0^{\circ})

cos(5x+40)= 3/5

cos (5 x + 4 0^{\circ}) = \frac{3}{5}

cos(4x)+cos(2x)+1=0

cos (4 x) + cos (2 x) + 1 = 0