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人気のある三角関数 >

4cos^2(x)-2=0

解

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

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

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

置換で解く

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

仮定:

cos (x) = u

4 u^{2} - 2 = 0

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

4 u^{2} - 2 = 0

2

を右側に移動します

4 u^{2} - 2 = 0

両辺に

2

を足す

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

簡素化

4 u^{2} = 2

4 u^{2} = 2

以下で両辺を割る

4

4 u^{2} = 2

以下で両辺を割る

4

\frac{4 u ^{2}}{4} = \frac{2}{4}

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

tan(3x)=sqrt(3)

tan (3 x) = 3

2 sin (x) = 0

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

cos(x)= 1/2 ,0<= x<= 2pi

cos (x) = \frac{1}{2}, 0 \leq x \leq 2 π

sec (θ) = 1