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受欢迎的三角函数 >

-9cos^2(θ)+13=20sin(θ)

解答

- 9 cos^{2} (θ) + 13 = 20 sin (θ)

解答

θ = 0.22409 \dots + 2 πn, θ = π - 0.22409 \dots + 2 πn

+1

度数

θ = 12.83958 \dots^{\circ} + 36 0^{\circ} n, θ = 167.16041 \dots^{\circ} + 36 0^{\circ} n

求解步骤

- 9 cos^{2} (θ) + 13 = 20 sin (θ)

两边减去

20 sin (θ)

- 9 cos^{2} (θ) + 13 - 20 sin (θ) = 0

使用三角恒等式改写

13 - 20 sin (θ) - 9 cos^{2} (θ)

使用毕达哥拉斯恒等式:

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

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

= 13 - 20 sin (θ) - 9 (1 - sin^{2} (θ))

化简

13 - 20 sin (θ) - 9 (1 - sin^{2} (θ)) : 9 sin^{2} (θ) - 20 sin (θ) + 4

13 - 20 sin (θ) - 9 (1 - sin^{2} (θ))

乘开

- 9 (1 - sin^{2} (θ)) : - 9 + 9 sin^{2} (θ)

- 9 (1 - sin^{2} (θ))

使用分配律:

a (b - c) = ab - a c

a = - 9, b = 1, c = sin^{2} (θ)

= - 9 \cdot 1 - (- 9) sin^{2} (θ)

使用加减运算法则

- (- a) = a

= - 9 \cdot 1 + 9 sin^{2} (θ)

数字相乘:

9 \cdot 1 = 9

= - 9 + 9 sin^{2} (θ)

= 13 - 20 sin (θ) - 9 + 9 sin^{2} (θ)

化简

13 - 20 sin (θ) - 9 + 9 sin^{2} (θ) : 9 sin^{2} (θ) - 20 sin (θ) + 4

13 - 20 sin (θ) - 9 + 9 sin^{2} (θ)

对同类项分组

= - 20 sin (θ) + 9 sin^{2} (θ) + 13 - 9

数字相加/相减:

13 - 9 = 4

= 9 sin^{2} (θ) - 20 sin (θ) + 4

= 9 sin^{2} (θ) - 20 sin (θ) + 4

= 9 sin^{2} (θ) - 20 sin (θ) + 4

4 - 20 sin (θ) + 9 sin^{2} (θ) = 0

用替代法求解

4 - 20 sin (θ) + 9 sin^{2} (θ) = 0

令

: sin (θ) = u

4 - 20 u + 9 u^{2} = 0

4 - 20 u + 9 u^{2} = 0 : u = 2, u = \frac{2}{9}

4 - 20 u + 9 u^{2} = 0

改写成标准形式

a x^{2} + b x + c = 0

9 u^{2} - 20 u + 4 = 0

使用求根公式求解

9 u^{2} - 20 u + 4 = 0

二次方程求根公式:

若

a = 9, b = - 20, c = 4

u_{1, 2} = \frac{- ( - 20 ) \pm ( - 20 ) ^{2} - 4 \cdot 9 \cdot 4}{2 \cdot 9}

u_{1, 2} = \frac{- ( - 20 ) \pm ( - 20 ) ^{2} - 4 \cdot 9 \cdot 4}{2 \cdot 9}

(- 20)^{2} - 4 \cdot 9 \cdot 4 = 16

(- 20)^{2} - 4 \cdot 9 \cdot 4

使用指数法则:

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

若

n

是偶数

(- 20)^{2} = 2 0^{2}

= 2 0^{2} - 4 \cdot 9 \cdot 4

数字相乘:

4 \cdot 9 \cdot 4 = 144

= 2 0^{2} - 144

2 0^{2} = 400

= 400 - 144

数字相减:

400 - 144 = 256

= 256

因式分解数字:

256 = 1 6^{2}

= 1 6^{2}

使用根式运算法则:

n a^{n} = a

1 6^{2} = 16

= 16

u_{1, 2} = \frac{- ( - 20 ) \pm 16}{2 \cdot 9}

将解分隔开

u_{1} = \frac{- ( - 20 ) + 16}{2 \cdot 9}, u_{2} = \frac{- ( - 20 ) - 16}{2 \cdot 9}

u = \frac{- ( - 20 ) + 16}{2 \cdot 9} : 2

\frac{- ( - 20 ) + 16}{2 \cdot 9}

使用法则

- (- a) = a

= \frac{20 + 16}{2 \cdot 9}

数字相加:

20 + 16 = 36

= \frac{36}{2 \cdot 9}

数字相乘:

2 \cdot 9 = 18

= \frac{36}{18}

数字相除:

\frac{36}{18} = 2

= 2

u = \frac{- ( - 20 ) - 16}{2 \cdot 9} : \frac{2}{9}

\frac{- ( - 20 ) - 16}{2 \cdot 9}

使用法则

- (- a) = a

= \frac{20 - 16}{2 \cdot 9}

数字相减:

20 - 16 = 4

= \frac{4}{2 \cdot 9}

数字相乘:

2 \cdot 9 = 18

= \frac{4}{18}

约分:

2

= \frac{2}{9}

二次方程组的解是:

u = 2, u = \frac{2}{9}

u = sin (θ)

代回

sin (θ) = 2, sin (θ) = \frac{2}{9}

sin (θ) = 2, sin (θ) = \frac{2}{9}

sin (θ) = 2 :

无解

sin (θ) = 2

- 1 \leq sin (x) \leq 1

无解

sin (θ) = \frac{2}{9} : θ = arcsin (\frac{2}{9}) + 2 πn, θ = π - arcsin (\frac{2}{9}) + 2 πn

sin (θ) = \frac{2}{9}

使用反三角函数性质

sin (θ) = \frac{2}{9}

sin (θ) = \frac{2}{9}

的通解

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

θ = arcsin (\frac{2}{9}) + 2 πn, θ = π - arcsin (\frac{2}{9}) + 2 πn

θ = arcsin (\frac{2}{9}) + 2 πn, θ = π - arcsin (\frac{2}{9}) + 2 πn

合并所有解

θ = arcsin (\frac{2}{9}) + 2 πn, θ = π - arcsin (\frac{2}{9}) + 2 πn

以小数形式表示解

θ = 0.22409 \dots + 2 πn, θ = π - 0.22409 \dots + 2 πn

作图

流行的例子

0 = sin (3 θ)

cos(x)=1,0<= x<2pi

cos (x) = 1, 0 \leq x < 2 π

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

5 csc (x) - 3 = 2

2 cos^{2} (x) + 2 = 4