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人気のある三角関数 >

11sin^2(θ)-5sin(θ)=4

解

11 sin^{2} (θ) - 5 sin (θ) = 4

解

θ = 1.05866 \dots + 2 πn, θ = π - 1.05866 \dots + 2 πn, θ = - 0.43031 \dots + 2 πn, θ = π + 0.43031 \dots + 2 πn

+1

度

θ = 60.65704 \dots^{\circ} + 36 0^{\circ} n, θ = 119.34295 \dots^{\circ} + 36 0^{\circ} n, θ = - 24.65520 \dots^{\circ} + 36 0^{\circ} n, θ = 204.65520 \dots^{\circ} + 36 0^{\circ} n

解答ステップ

11 sin^{2} (θ) - 5 sin (θ) = 4

置換で解く

11 sin^{2} (θ) - 5 sin (θ) = 4

仮定:

sin (θ) = u

11 u^{2} - 5 u = 4

11 u^{2} - 5 u = 4 : u = \frac{5 + 201}{22}, u = \frac{5 - 201}{22}

11 u^{2} - 5 u = 4

4

を左側に移動します

11 u^{2} - 5 u = 4

両辺から

4

を引く

11 u^{2} - 5 u - 4 = 4 - 4

簡素化

11 u^{2} - 5 u - 4 = 0

11 u^{2} - 5 u - 4 = 0

解くとthe二次式

11 u^{2} - 5 u - 4 = 0

二次Equationの公式:

次の場合:

a = 11, b = - 5, c = - 4

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 11 ( - 4 )}{2 \cdot 11}

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 11 ( - 4 )}{2 \cdot 11}

(- 5)^{2} - 4 \cdot 11 (- 4) = 201

(- 5)^{2} - 4 \cdot 11 (- 4)

規則を適用

- (- a) = a

= (- 5)^{2} + 4 \cdot 11 \cdot 4

指数の規則を適用する:

n

が偶数であれば

(- a)^{n} = a^{n}

(- 5)^{2} = 5^{2}

= 5^{2} + 4 \cdot 11 \cdot 4

数を乗じる:

4 \cdot 11 \cdot 4 = 176

= 5^{2} + 176

5^{2} = 25

= 25 + 176

数を足す:

25 + 176 = 201

= 201

u_{1, 2} = \frac{- ( - 5 ) \pm 201}{2 \cdot 11}

解を分離する

u_{1} = \frac{- ( - 5 ) + 201}{2 \cdot 11}, u_{2} = \frac{- ( - 5 ) - 201}{2 \cdot 11}

u = \frac{- ( - 5 ) + 201}{2 \cdot 11} : \frac{5 + 201}{22}

\frac{- ( - 5 ) + 201}{2 \cdot 11}

規則を適用

- (- a) = a

= \frac{5 + 201}{2 \cdot 11}

数を乗じる:

2 \cdot 11 = 22

= \frac{5 + 201}{22}

u = \frac{- ( - 5 ) - 201}{2 \cdot 11} : \frac{5 - 201}{22}

\frac{- ( - 5 ) - 201}{2 \cdot 11}

規則を適用

- (- a) = a

= \frac{5 - 201}{2 \cdot 11}

数を乗じる:

2 \cdot 11 = 22

= \frac{5 - 201}{22}

二次equationの解:

u = \frac{5 + 201}{22}, u = \frac{5 - 201}{22}

代用を戻す

u = sin (θ)

sin (θ) = \frac{5 + 201}{22}, sin (θ) = \frac{5 - 201}{22}

sin (θ) = \frac{5 + 201}{22}, sin (θ) = \frac{5 - 201}{22}

sin (θ) = \frac{5 + 201}{22} : θ = arcsin (\frac{5 + 201}{22}) + 2 πn, θ = π - arcsin (\frac{5 + 201}{22}) + 2 πn

sin (θ) = \frac{5 + 201}{22}

三角関数の逆数プロパティを適用する

sin (θ) = \frac{5 + 201}{22}

以下の一般解

sin (θ) = \frac{5 + 201}{22}

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

θ = arcsin (\frac{5 + 201}{22}) + 2 πn, θ = π - arcsin (\frac{5 + 201}{22}) + 2 πn

θ = arcsin (\frac{5 + 201}{22}) + 2 πn, θ = π - arcsin (\frac{5 + 201}{22}) + 2 πn

sin (θ) = \frac{5 - 201}{22} : θ = arcsin (\frac{5 - 201}{22}) + 2 πn, θ = π + arcsin (- \frac{5 - 201}{22}) + 2 πn

sin (θ) = \frac{5 - 201}{22}

三角関数の逆数プロパティを適用する

sin (θ) = \frac{5 - 201}{22}

以下の一般解

sin (θ) = \frac{5 - 201}{22}

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

θ = arcsin (\frac{5 - 201}{22}) + 2 πn, θ = π + arcsin (- \frac{5 - 201}{22}) + 2 πn

θ = arcsin (\frac{5 - 201}{22}) + 2 πn, θ = π + arcsin (- \frac{5 - 201}{22}) + 2 πn

すべての解を組み合わせる

θ = arcsin (\frac{5 + 201}{22}) + 2 πn, θ = π - arcsin (\frac{5 + 201}{22}) + 2 πn, θ = arcsin (\frac{5 - 201}{22}) + 2 πn, θ = π + arcsin (- \frac{5 - 201}{22}) + 2 πn

10進法形式で解を証明する

θ = 1.05866 \dots + 2 πn, θ = π - 1.05866 \dots + 2 πn, θ = - 0.43031 \dots + 2 πn, θ = π + 0.43031 \dots + 2 πn

グラフ

人気の例

5cos(x)cot(x)=1

5 cos (x) cot (x) = 1

- 0.5 = cos (3 x)

sin (3 x) = - 0.5

tan (x) = \frac{5}{8}

cos (x) = \frac{7}{8}