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人気のある三角関数 >

cos^2(x)= 1/3

解

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

解

x = 0.95531 \dots + 2 πn, x = 2 π - 0.95531 \dots + 2 πn, x = 2.18627 \dots + 2 πn, x = - 2.18627 \dots + 2 πn

+1

度

x = 54.73561 \dots^{\circ} + 36 0^{\circ} n, x = 305.26438 \dots^{\circ} + 36 0^{\circ} n, x = 125.26438 \dots^{\circ} + 36 0^{\circ} n, x = - 125.26438 \dots^{\circ} + 36 0^{\circ} n

解答ステップ

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

置換で解く

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

仮定:

cos (x) = u

u^{2} = \frac{1}{3}

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

u^{2} = \frac{1}{3}

x^{2} = f (a)

の場合, 解は

x = f (a), - f (a)

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

代用を戻す

u = cos (x)

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

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

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

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

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

10進法形式で解を証明する

x = 0.95531 \dots + 2 πn, x = 2 π - 0.95531 \dots + 2 πn, x = 2.18627 \dots + 2 πn, x = - 2.18627 \dots + 2 πn

グラフ

人気の例

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tan(θ)= 3/(sqrt(3))

tan (θ) = \frac{3}{3}

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