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受欢迎的三角函数 >

-3cos^2(θ)+11=-5sin(θ)+6

解答

- 3 cos^{2} (θ) + 11 = - 5 sin (θ) + 6

解答

θ = - 0.72972 \dots + 2 πn, θ = π + 0.72972 \dots + 2 πn, θ = \frac{3 π}{2} + 2 πn

+1

度数

θ = - 41.81031 \dots^{\circ} + 36 0^{\circ} n, θ = 221.81031 \dots^{\circ} + 36 0^{\circ} n, θ = 27 0^{\circ} + 36 0^{\circ} n

求解步骤

- 3 cos^{2} (θ) + 11 = - 5 sin (θ) + 6

两边减去

- 5 sin (θ) + 6

5 sin (θ) - 3 cos^{2} (θ) + 5 = 0

使用三角恒等式改写

5 - 3 cos^{2} (θ) + 5 sin (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)

= 5 - 3 (1 - sin^{2} (θ)) + 5 sin (θ)

化简

5 - 3 (1 - sin^{2} (θ)) + 5 sin (θ) : 3 sin^{2} (θ) + 5 sin (θ) + 2

5 - 3 (1 - sin^{2} (θ)) + 5 sin (θ)

乘开

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

- 3 (1 - sin^{2} (θ))

使用分配律:

a (b - c) = ab - a c

a = - 3, b = 1, c = sin^{2} (θ)

= - 3 \cdot 1 - (- 3) sin^{2} (θ)

使用加减运算法则

- (- a) = a

= - 3 \cdot 1 + 3 sin^{2} (θ)

数字相乘:

3 \cdot 1 = 3

= - 3 + 3 sin^{2} (θ)

= 5 - 3 + 3 sin^{2} (θ) + 5 sin (θ)

数字相减:

5 - 3 = 2

= 3 sin^{2} (θ) + 5 sin (θ) + 2

= 3 sin^{2} (θ) + 5 sin (θ) + 2

2 + 3 sin^{2} (θ) + 5 sin (θ) = 0

用替代法求解

2 + 3 sin^{2} (θ) + 5 sin (θ) = 0

令

: sin (θ) = u

2 + 3 u^{2} + 5 u = 0

2 + 3 u^{2} + 5 u = 0 : u = - \frac{2}{3}, u = - 1

2 + 3 u^{2} + 5 u = 0

改写成标准形式

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

3 u^{2} + 5 u + 2 = 0

使用求根公式求解

3 u^{2} + 5 u + 2 = 0

二次方程求根公式:

若

a = 3, b = 5, c = 2

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 3 \cdot 2}{2 \cdot 3}

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 3 \cdot 2}{2 \cdot 3}

5^{2} - 4 \cdot 3 \cdot 2 = 1

5^{2} - 4 \cdot 3 \cdot 2

数字相乘:

4 \cdot 3 \cdot 2 = 24

= 5^{2} - 24

5^{2} = 25

= 25 - 24

数字相减:

25 - 24 = 1

= 1

使用法则

1 = 1

= 1

u_{1, 2} = \frac{- 5 \pm 1}{2 \cdot 3}

将解分隔开

u_{1} = \frac{- 5 + 1}{2 \cdot 3}, u_{2} = \frac{- 5 - 1}{2 \cdot 3}

u = \frac{- 5 + 1}{2 \cdot 3} : - \frac{2}{3}

\frac{- 5 + 1}{2 \cdot 3}

数字相加/相减:

- 5 + 1 = - 4

= \frac{- 4}{2 \cdot 3}

数字相乘:

2 \cdot 3 = 6

= \frac{- 4}{6}

使用分式法则:

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

= - \frac{4}{6}

约分:

2

= - \frac{2}{3}

u = \frac{- 5 - 1}{2 \cdot 3} : - 1

\frac{- 5 - 1}{2 \cdot 3}

数字相减:

- 5 - 1 = - 6

= \frac{- 6}{2 \cdot 3}

数字相乘:

2 \cdot 3 = 6

= \frac{- 6}{6}

使用分式法则:

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

= - \frac{6}{6}

使用法则

\frac{a}{a} = 1

= - 1

二次方程组的解是:

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

u = sin (θ)

代回

sin (θ) = - \frac{2}{3}, sin (θ) = - 1

sin (θ) = - \frac{2}{3}, sin (θ) = - 1

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

sin (θ) = - \frac{2}{3}

使用反三角函数性质

sin (θ) = - \frac{2}{3}

sin (θ) = - \frac{2}{3}

的通解

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

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

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

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

sin (θ) = - 1

sin (θ) = - 1

的通解

sin (x)

周期表（周期为

2 πn

"）：

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

θ = \frac{3 π}{2} + 2 πn

θ = \frac{3 π}{2} + 2 πn

合并所有解

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

以小数形式表示解

θ = - 0.72972 \dots + 2 πn, θ = π + 0.72972 \dots + 2 πn, θ = \frac{3 π}{2} + 2 πn

作图

流行的例子

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cos(2x)=-sin(x)

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