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人気のある三角関数 >

2cos^2(x)-1=0,0<= x<= 2pi

解

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

解

x = \frac{π}{4}, x = \frac{7 π}{4}, x = \frac{3 π}{4}, x = \frac{5 π}{4}

+1

度

x = 4 5^{\circ}, x = 31 5^{\circ}, x = 13 5^{\circ}, x = 22 5^{\circ}

解答ステップ

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

置換で解く

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

仮定:

cos (x) = u

2 u^{2} - 1 = 0

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

2 u^{2} - 1 = 0

1

を右側に移動します

2 u^{2} - 1 = 0

両辺に

1

を足す

2 u^{2} - 1 + 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}, 0 \leq x \leq 2 π : x = \frac{π}{4}, x = \frac{7 π}{4}

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

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

範囲の解答

0 \leq x \leq 2 π

x = \frac{π}{4}, x = \frac{7 π}{4}

cos (x) = - \frac{1}{2}, 0 \leq x \leq 2 π : x = \frac{3 π}{4}, x = \frac{5 π}{4}

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

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

範囲の解答

0 \leq x \leq 2 π

x = \frac{3 π}{4}, x = \frac{5 π}{4}

すべての解を組み合わせる

x = \frac{π}{4}, x = \frac{7 π}{4}, x = \frac{3 π}{4}, x = \frac{5 π}{4}

グラフ

人気の例

sin (θ) = 0.82

tan (x) = \frac{1}{10}

sqrt(cos^2(2x)+1)+sin(2x)=2

cos^{2} (2 x) + 1 + sin (2 x) = 2

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