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人気のある三角関数 >

solvefor x,cos^2(x)+cos^2(y)cos^2(z)=1

解

解く

x, cos^{2} (x) + cos^{2} (y) cos^{2} (z) = 1

解

x = arcsin (- cos (y) cos (z)) + 2 πn, x = π + arcsin (cos (y) cos (z)) + 2 πn, x = arcsin (cos (y) cos (z)) + 2 πn, x = π + arcsin (- cos (y) cos (z)) + 2 πn

解答ステップ

cos^{2} (x) + cos^{2} (y) cos^{2} (z) = 1

両辺から

1

を引く

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

三角関数の公式を使用して書き換える

- 1 + cos^{2} (x) + cos^{2} (y) cos^{2} (z)

ピタゴラスの公式を使用する:

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

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

= cos^{2} (y) cos^{2} (z) - sin^{2} (x)

- sin^{2} (x) + cos^{2} (y) cos^{2} (z) = 0

因数

- sin^{2} (x) + cos^{2} (y) cos^{2} (z) : (cos (y) cos (z) + sin (x)) (cos (y) cos (z) - sin (x))

- sin^{2} (x) + cos^{2} (y) cos^{2} (z)

cos^{2} (y) cos^{2} (z)

を書き換え

(cos (y) cos (z))^{2}

cos^{2} (y) cos^{2} (z)

指数の規則を適用する:

a^{m} b^{m} = (ab)^{m}

cos^{2} (y) cos^{2} (z) = (cos (y) cos (z))^{2}

= (cos (y) cos (z))^{2}

= - sin^{2} (x) + (cos (y) cos (z))^{2}

2乗の差の公式を適用する:

x^{2} - y^{2} = (x + y) (x - y)

(cos (y) cos (z))^{2} - sin^{2} (x) = (cos (y) cos (z) + sin (x)) (cos (y) cos (z) - sin (x))

= (cos (y) cos (z) + sin (x)) (cos (y) cos (z) - sin (x))

(cos (y) cos (z) + sin (x)) (cos (y) cos (z) - sin (x)) = 0

各部分を別個に解く

cos (y) cos (z) + sin (x) = 0 or cos (y) cos (z) - sin (x) = 0

cos (y) cos (z) + sin (x) = 0 : x = arcsin (- cos (y) cos (z)) + 2 πn, x = π + arcsin (cos (y) cos (z)) + 2 πn

cos (y) cos (z) + sin (x) = 0

cos (y) cos (z)

を右側に移動します

cos (y) cos (z) + sin (x) = 0

両辺から

cos (y) cos (z)

を引く

cos (y) cos (z) + sin (x) - cos (y) cos (z) = 0 - cos (y) cos (z)

簡素化

sin (x) = - cos (y) cos (z)

sin (x) = - cos (y) cos (z)

三角関数の逆数プロパティを適用する

sin (x) = - cos (y) cos (z)

以下の一般解

sin (x) = - cos (y) cos (z)

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

x = arcsin (- cos (y) cos (z)) + 2 πn, x = π + arcsin (cos (y) cos (z)) + 2 πn

x = arcsin (- cos (y) cos (z)) + 2 πn, x = π + arcsin (cos (y) cos (z)) + 2 πn

cos (y) cos (z) - sin (x) = 0 : x = arcsin (cos (y) cos (z)) + 2 πn, x = π + arcsin (- cos (y) cos (z)) + 2 πn

cos (y) cos (z) - sin (x) = 0

cos (y) cos (z)

を右側に移動します

cos (y) cos (z) - sin (x) = 0

両辺から

cos (y) cos (z)

を引く

cos (y) cos (z) - sin (x) - cos (y) cos (z) = 0 - cos (y) cos (z)

簡素化

- sin (x) = - cos (y) cos (z)

- sin (x) = - cos (y) cos (z)

以下で両辺を割る

- 1

- sin (x) = - cos (y) cos (z)

以下で両辺を割る

- 1

\frac{- sin ( x )}{- 1} = \frac{- cos ( y ) cos ( z )}{- 1}

簡素化

sin (x) = cos (y) cos (z)

sin (x) = cos (y) cos (z)

三角関数の逆数プロパティを適用する

sin (x) = cos (y) cos (z)

以下の一般解

sin (x) = cos (y) cos (z)

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

x = arcsin (cos (y) cos (z)) + 2 πn, x = π + arcsin (- cos (y) cos (z)) + 2 πn

x = arcsin (cos (y) cos (z)) + 2 πn, x = π + arcsin (- cos (y) cos (z)) + 2 πn

すべての解を組み合わせる

x = arcsin (- cos (y) cos (z)) + 2 πn, x = π + arcsin (cos (y) cos (z)) + 2 πn, x = arcsin (cos (y) cos (z)) + 2 πn, x = π + arcsin (- cos (y) cos (z)) + 2 πn

グラフ

人気の例

cos(5x)+cos(x)=0

cos (5 x) + cos (x) = 0

solvefor x,f=cos^2(x)

so l v e f or x, f = cos^{2} (x)

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

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

5cos^2(x)+3cos(x)-2=0

5 cos^{2} (x) + 3 cos (x) - 2 = 0

sin(3x)=sin(6x)

sin (3 x) = sin (6 x)