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人気のある三角関数 >

2cos^2(x)-sin^2(x)=1

解

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

解

x = 0.61547 \dots + 2 πn, x = π - 0.61547 \dots + 2 πn, x = - 0.61547 \dots + 2 πn, x = π + 0.61547 \dots + 2 πn

+1

度

x = 35.26438 \dots^{\circ} + 36 0^{\circ} n, x = 144.73561 \dots^{\circ} + 36 0^{\circ} n, x = - 35.26438 \dots^{\circ} + 36 0^{\circ} n, x = 215.26438 \dots^{\circ} + 36 0^{\circ} n

解答ステップ

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

両辺から

1

を引く

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

三角関数の公式を使用して書き換える

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

ピタゴラスの公式を使用する:

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

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

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

簡素化

- 1 - sin^{2} (x) + 2 (1 - sin^{2} (x)) : - 3 sin^{2} (x) + 1

- 1 - sin^{2} (x) + 2 (1 - sin^{2} (x))

拡張

2 (1 - sin^{2} (x)) : 2 - 2 sin^{2} (x)

2 (1 - sin^{2} (x))

分配法則を適用する:

a (b - c) = ab - a c

a = 2, b = 1, c = sin^{2} (x)

= 2 \cdot 1 - 2 sin^{2} (x)

数を乗じる:

2 \cdot 1 = 2

= 2 - 2 sin^{2} (x)

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

簡素化

- 1 - sin^{2} (x) + 2 - 2 sin^{2} (x) : - 3 sin^{2} (x) + 1

- 1 - sin^{2} (x) + 2 - 2 sin^{2} (x)

条件のようなグループ

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

類似した元を足す:

- sin^{2} (x) - 2 sin^{2} (x) = - 3 sin^{2} (x)

= - 3 sin^{2} (x) - 1 + 2

数を足す/引く:

- 1 + 2 = 1

= - 3 sin^{2} (x) + 1

= - 3 sin^{2} (x) + 1

= - 3 sin^{2} (x) + 1

1 - 3 sin^{2} (x) = 0

置換で解く

1 - 3 sin^{2} (x) = 0

仮定:

sin (x) = u

1 - 3 u^{2} = 0

1 - 3 u^{2} = 0 : u = \frac{1}{3}, u = - \frac{1}{3}

1 - 3 u^{2} = 0

1

を右側に移動します

1 - 3 u^{2} = 0

両辺から

1

を引く

1 - 3 u^{2} - 1 = 0 - 1

簡素化

- 3 u^{2} = - 1

- 3 u^{2} = - 1

以下で両辺を割る

- 3

- 3 u^{2} = - 1

以下で両辺を割る

- 3

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

簡素化

u^{2} = \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 = sin (x)

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

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

sin (x) = \frac{1}{3} : x = arcsin (\frac{1}{3}) + 2 πn, x = π - arcsin (\frac{1}{3}) + 2 πn

sin (x) = \frac{1}{3}

三角関数の逆数プロパティを適用する

sin (x) = \frac{1}{3}

以下の一般解

sin (x) = \frac{1}{3}

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

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

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

sin (x) = - \frac{1}{3} : x = arcsin (- \frac{1}{3}) + 2 πn, x = π + arcsin (\frac{1}{3}) + 2 πn

sin (x) = - \frac{1}{3}

三角関数の逆数プロパティを適用する

sin (x) = - \frac{1}{3}

以下の一般解

sin (x) = - \frac{1}{3}

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

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

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

すべての解を組み合わせる

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

10進法形式で解を証明する

x = 0.61547 \dots + 2 πn, x = π - 0.61547 \dots + 2 πn, x = - 0.61547 \dots + 2 πn, x = π + 0.61547 \dots + 2 πn

グラフ

人気の例

2cos^2(θ)-sin(θ)-1=0

2 cos^{2} (θ) - sin (θ) - 1 = 0

sin (t) = - 1

sin(3x)=1,0<= x<= 2pi

sin (3 x) = 1, 0 \leq x \leq 2 π

solvefor x,z=arcsin(x/(sqrt(x^2+y^2)))

so l v e f or x, z = arcsin (\frac{x}{x ^{2} + y ^{2}})

cos(θ)(2cos(θ)+1)=0

cos (θ) (2 cos (θ) + 1) = 0