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人気のある三角関数 >

16sin^2(x)-1=0

解

16 sin^{2} (x) - 1 = 0

解

x = 0.25268 \dots + 2 πn, x = π - 0.25268 \dots + 2 πn, x = - 0.25268 \dots + 2 πn, x = π + 0.25268 \dots + 2 πn

+1

度

x = 14.47751 \dots^{\circ} + 36 0^{\circ} n, x = 165.52248 \dots^{\circ} + 36 0^{\circ} n, x = - 14.47751 \dots^{\circ} + 36 0^{\circ} n, x = 194.47751 \dots^{\circ} + 36 0^{\circ} n

解答ステップ

16 sin^{2} (x) - 1 = 0

置換で解く

16 sin^{2} (x) - 1 = 0

仮定:

sin (x) = u

16 u^{2} - 1 = 0

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

16 u^{2} - 1 = 0

1

を右側に移動します

16 u^{2} - 1 = 0

両辺に

1

を足す

16 u^{2} - 1 + 1 = 0 + 1

簡素化

16 u^{2} = 1

16 u^{2} = 1

以下で両辺を割る

16

16 u^{2} = 1

以下で両辺を割る

16

\frac{16 u ^{2}}{16} = \frac{1}{16}

簡素化

u^{2} = \frac{1}{16}

u^{2} = \frac{1}{16}

x^{2} = f (a)

の場合, 解は

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

u = \frac{1}{16}, u = - \frac{1}{16}

\frac{1}{16} = \frac{1}{4}

\frac{1}{16}

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{1}{16}

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{1}{4}

規則を適用

1 = 1

= \frac{1}{4}

- \frac{1}{16} = - \frac{1}{4}

- \frac{1}{16}

簡素化

\frac{1}{16} : \frac{1}{4}

\frac{1}{16}

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{1}{16}

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{1}{4}

= - \frac{1}{4}

規則を適用

1 = 1

= - \frac{1}{4}

u = \frac{1}{4}, u = - \frac{1}{4}

代用を戻す

u = sin (x)

sin (x) = \frac{1}{4}, sin (x) = - \frac{1}{4}

sin (x) = \frac{1}{4}, sin (x) = - \frac{1}{4}

sin (x) = \frac{1}{4} : x = arcsin (\frac{1}{4}) + 2 πn, x = π - arcsin (\frac{1}{4}) + 2 πn

sin (x) = \frac{1}{4}

三角関数の逆数プロパティを適用する

sin (x) = \frac{1}{4}

以下の一般解

sin (x) = \frac{1}{4}

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

x = arcsin (\frac{1}{4}) + 2 πn, x = π - arcsin (\frac{1}{4}) + 2 πn

x = arcsin (\frac{1}{4}) + 2 πn, x = π - arcsin (\frac{1}{4}) + 2 πn

sin (x) = - \frac{1}{4} : x = arcsin (- \frac{1}{4}) + 2 πn, x = π + arcsin (\frac{1}{4}) + 2 πn

sin (x) = - \frac{1}{4}

三角関数の逆数プロパティを適用する

sin (x) = - \frac{1}{4}

以下の一般解

sin (x) = - \frac{1}{4}

sin (x) = - a \Rightarrow x = arcsin (- a) + 2 πn, x = π + arcsin (a) + 2 πn

x = arcsin (- \frac{1}{4}) + 2 πn, x = π + arcsin (\frac{1}{4}) + 2 πn

x = arcsin (- \frac{1}{4}) + 2 πn, x = π + arcsin (\frac{1}{4}) + 2 πn

すべての解を組み合わせる

x = arcsin (\frac{1}{4}) + 2 πn, x = π - arcsin (\frac{1}{4}) + 2 πn, x = arcsin (- \frac{1}{4}) + 2 πn, x = π + arcsin (\frac{1}{4}) + 2 πn

10進法形式で解を証明する

x = 0.25268 \dots + 2 πn, x = π - 0.25268 \dots + 2 πn, x = - 0.25268 \dots + 2 πn, x = π + 0.25268 \dots + 2 πn

グラフ

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