ja

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

人気のある三角関数 >

solvefor t,x=acos^2(t)+bsin^2(t)

解

解く

t, x = a cos^{2} (t) + b sin^{2} (t)

解

t = arcsin (\frac{x - a}{- a + b}) + 2 πn, t = π + arcsin (- \frac{x - a}{- a + b}) + 2 πn, t = arcsin (- \frac{x - a}{- a + b}) + 2 πn, t = π + arcsin (\frac{x - a}{- a + b}) + 2 πn

解答ステップ

x = a cos^{2} (t) + b sin^{2} (t)

辺を交換する

a cos^{2} (t) + b sin^{2} (t) = x

両辺から

x

を引く

a cos^{2} (t) + b sin^{2} (t) - x = 0

三角関数の公式を使用して書き換える

- x + cos^{2} (t) a + sin^{2} (t) b

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)

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

- x + (1 - sin^{2} (t)) a + sin^{2} (t) b = 0

置換で解く

- x + (1 - sin^{2} (t)) a + sin^{2} (t) b = 0

仮定:

sin (t) = u

- x + (1 - u^{2}) a + u^{2} b = 0

- x + (1 - u^{2}) a + u^{2} b = 0 : u = \frac{x - a}{- a + b}, u = - \frac{x - a}{- a + b}; - a + b \neq = 0

- x + (1 - u^{2}) a + u^{2} b = 0

x

を右側に移動します

- x + (1 - u^{2}) a + u^{2} b = 0

両辺に

x

を足す

- x + (1 - u^{2}) a + u^{2} b + x = 0 + x

簡素化

(1 - u^{2}) a + u^{2} b = x

(1 - u^{2}) a + u^{2} b = x

拡張

(1 - u^{2}) a + u^{2} b : a - a u^{2} + b u^{2}

(1 - u^{2}) a + u^{2} b

= a (1 - u^{2}) + b u^{2}

拡張

a (1 - u^{2}) : a - a u^{2}

a (1 - u^{2})

分配法則を適用する:

a (b - c) = ab - a c

a = a, b = 1, c = u^{2}

= a \cdot 1 - a u^{2}

= 1 \cdot a - a u^{2}

乗算:

1 \cdot a = a

= a - a u^{2}

= a - a u^{2} + u^{2} b

= a - a u^{2} + b u^{2}

a - a u^{2} + b u^{2} = x

a

を右側に移動します

a - a u^{2} + b u^{2} = x

両辺から

a

を引く

a - a u^{2} + b u^{2} - a = x - a

簡素化

- a u^{2} + b u^{2} = x - a

- a u^{2} + b u^{2} = x - a

因数

- a u^{2} + b u^{2} : u^{2} (- a + b)

- a u^{2} + b u^{2}

共通項をくくり出す

u^{2}

= u^{2} (- a + b)

u^{2} (- a + b) = x - a

以下で両辺を割る

- a + b; - a + b \neq = 0

u^{2} (- a + b) = x - a

以下で両辺を割る

- a + b; - a + b \neq = 0

\frac{u ^{2} ( - a + b )}{- a + b} = \frac{x}{- a + b} - \frac{a}{- a + b}; - a + b \neq = 0

簡素化

\frac{u ^{2} ( - a + b )}{- a + b} = \frac{x}{- a + b} - \frac{a}{- a + b}

簡素化

\frac{u ^{2} ( - a + b )}{- a + b} : u^{2}

\frac{u ^{2} ( - a + b )}{- a + b}

共通因数を約分する:

- a + b

= u^{2}

簡素化

\frac{x}{- a + b} - \frac{a}{- a + b} : \frac{x - a}{- a + b}

\frac{x}{- a + b} - \frac{a}{- a + b}

規則を適用

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{x - a}{- a + b}

u^{2} = \frac{x - a}{- a + b}; - a + b \neq = 0

u^{2} = \frac{x - a}{- a + b}; - a + b \neq = 0

u^{2} = \frac{x - a}{- a + b}; - a + b \neq = 0

x^{2} = f (a)

の場合, 解は

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

u = \frac{x - a}{- a + b}, u = - \frac{x - a}{- a + b}; - a + b \neq = 0

代用を戻す

u = sin (t)

sin (t) = \frac{x - a}{- a + b}, sin (t) = - \frac{x - a}{- a + b}; - a + b \neq = 0

sin (t) = \frac{x - a}{- a + b}, sin (t) = - \frac{x - a}{- a + b}; - a + b \neq = 0

sin (t) = \frac{x - a}{- a + b} : t = arcsin (\frac{x - a}{- a + b}) + 2 πn, t = π + arcsin (- \frac{x - a}{- a + b}) + 2 πn

sin (t) = \frac{x - a}{- a + b}

三角関数の逆数プロパティを適用する

sin (t) = \frac{x - a}{- a + b}

以下の一般解

sin (t) = \frac{x - a}{- a + b}

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

t = arcsin (\frac{x - a}{- a + b}) + 2 πn, t = π + arcsin (- \frac{x - a}{- a + b}) + 2 πn

t = arcsin (\frac{x - a}{- a + b}) + 2 πn, t = π + arcsin (- \frac{x - a}{- a + b}) + 2 πn

sin (t) = - \frac{x - a}{- a + b} : t = arcsin (- \frac{x - a}{- a + b}) + 2 πn, t = π + arcsin (\frac{x - a}{- a + b}) + 2 πn

sin (t) = - \frac{x - a}{- a + b}

三角関数の逆数プロパティを適用する

sin (t) = - \frac{x - a}{- a + b}

以下の一般解

sin (t) = - \frac{x - a}{- a + b}

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

t = arcsin (- \frac{x - a}{- a + b}) + 2 πn, t = π + arcsin (\frac{x - a}{- a + b}) + 2 πn

t = arcsin (- \frac{x - a}{- a + b}) + 2 πn, t = π + arcsin (\frac{x - a}{- a + b}) + 2 πn

すべての解を組み合わせる

t = arcsin (\frac{x - a}{- a + b}) + 2 πn, t = π + arcsin (- \frac{x - a}{- a + b}) + 2 πn, t = arcsin (- \frac{x - a}{- a + b}) + 2 πn, t = π + arcsin (\frac{x - a}{- a + b}) + 2 πn

グラフ

人気の例

cos(3x)-sin(3x)=0

cos (3 x) - sin (3 x) = 0

8sin^2(x)cos^2(x)=1

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

sin(2x)= 1/2 ,-pi<= x<= pi

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

tan^2(θ)-3cot(θ)=0

tan^{2} (θ) - 3 cot (θ) = 0

solvefor a,cos(ax)+(b/x)sin(ax)=0

so l v e f or a, cos (a x) + (\frac{b}{x}) sin (a x) = 0