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人気のある三角関数 >

25cos^2(x)=16

解

25 cos^{2} (x) = 16

解

x = 0.64350 \dots + 2 πn, x = 2 π - 0.64350 \dots + 2 πn, x = 2.49809 \dots + 2 πn, x = - 2.49809 \dots + 2 πn

+1

度

x = 36.86989 \dots^{\circ} + 36 0^{\circ} n, x = 323.13010 \dots^{\circ} + 36 0^{\circ} n, x = 143.13010 \dots^{\circ} + 36 0^{\circ} n, x = - 143.13010 \dots^{\circ} + 36 0^{\circ} n

解答ステップ

25 cos^{2} (x) = 16

置換で解く

25 cos^{2} (x) = 16

仮定:

cos (x) = u

25 u^{2} = 16

25 u^{2} = 16 : u = \frac{4}{5}, u = - \frac{4}{5}

25 u^{2} = 16

以下で両辺を割る

25

25 u^{2} = 16

以下で両辺を割る

25

\frac{25 u ^{2}}{25} = \frac{16}{25}

簡素化

u^{2} = \frac{16}{25}

u^{2} = \frac{16}{25}

x^{2} = f (a)

の場合, 解は

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

u = \frac{16}{25}, u = - \frac{16}{25}

\frac{16}{25} = \frac{4}{5}

\frac{16}{25}

累乗根の規則を適用する:

n \frac{a}{b} = \frac{n a}{n b},

, 以下を想定

a \geq 0, b \geq 0

= \frac{16}{25}

25 = 5

25

数を因数に分解する:

25 = 5^{2}

= 5^{2}

累乗根の規則を適用する:

n a^{n} = a

5^{2} = 5

= 5

= \frac{16}{5}

16 = 4

16

数を因数に分解する:

16 = 4^{2}

= 4^{2}

累乗根の規則を適用する:

n a^{n} = a

4^{2} = 4

= 4

= \frac{4}{5}

- \frac{16}{25} = - \frac{4}{5}

- \frac{16}{25}

簡素化

\frac{16}{25} : \frac{4}{5}

\frac{16}{25}

累乗根の規則を適用する:

n \frac{a}{b} = \frac{n a}{n b},

, 以下を想定

a \geq 0, b \geq 0

= \frac{16}{25}

25 = 5

25

数を因数に分解する:

25 = 5^{2}

= 5^{2}

累乗根の規則を適用する:

n a^{n} = a

5^{2} = 5

= 5

= \frac{16}{5}

16 = 4

16

数を因数に分解する:

16 = 4^{2}

= 4^{2}

累乗根の規則を適用する:

n a^{n} = a

4^{2} = 4

= 4

= \frac{4}{5}

= - \frac{4}{5}

u = \frac{4}{5}, u = - \frac{4}{5}

代用を戻す

u = cos (x)

cos (x) = \frac{4}{5}, cos (x) = - \frac{4}{5}

cos (x) = \frac{4}{5}, cos (x) = - \frac{4}{5}

cos (x) = \frac{4}{5} : x = arccos (\frac{4}{5}) + 2 πn, x = 2 π - arccos (\frac{4}{5}) + 2 πn

cos (x) = \frac{4}{5}

三角関数の逆数プロパティを適用する

cos (x) = \frac{4}{5}

以下の一般解

cos (x) = \frac{4}{5}

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

x = arccos (\frac{4}{5}) + 2 πn, x = 2 π - arccos (\frac{4}{5}) + 2 πn

x = arccos (\frac{4}{5}) + 2 πn, x = 2 π - arccos (\frac{4}{5}) + 2 πn

cos (x) = - \frac{4}{5} : x = arccos (- \frac{4}{5}) + 2 πn, x = - arccos (- \frac{4}{5}) + 2 πn

cos (x) = - \frac{4}{5}

三角関数の逆数プロパティを適用する

cos (x) = - \frac{4}{5}

以下の一般解

cos (x) = - \frac{4}{5}

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

x = arccos (- \frac{4}{5}) + 2 πn, x = - arccos (- \frac{4}{5}) + 2 πn

x = arccos (- \frac{4}{5}) + 2 πn, x = - arccos (- \frac{4}{5}) + 2 πn

すべての解を組み合わせる

x = arccos (\frac{4}{5}) + 2 πn, x = 2 π - arccos (\frac{4}{5}) + 2 πn, x = arccos (- \frac{4}{5}) + 2 πn, x = - arccos (- \frac{4}{5}) + 2 πn

10進法形式で解を証明する

x = 0.64350 \dots + 2 πn, x = 2 π - 0.64350 \dots + 2 πn, x = 2.49809 \dots + 2 πn, x = - 2.49809 \dots + 2 πn

グラフ

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