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受欢迎的三角函数 >

-2cos^2(θ)-4sin(θ)+2=-5sin(θ)

解答

- 2 cos^{2} (θ) - 4 sin (θ) + 2 = - 5 sin (θ)

解答

θ = 2 πn, θ = π + 2 πn, θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

+1

度数

θ = 0^{\circ} + 36 0^{\circ} n, θ = 18 0^{\circ} + 36 0^{\circ} n, θ = 21 0^{\circ} + 36 0^{\circ} n, θ = 33 0^{\circ} + 36 0^{\circ} n

求解步骤

- 2 cos^{2} (θ) - 4 sin (θ) + 2 = - 5 sin (θ)

两边减去

- 5 sin (θ)

- 2 cos^{2} (θ) + sin (θ) + 2 = 0

使用三角恒等式改写

2 + sin (θ) - 2 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)

= 2 + sin (θ) - 2 (1 - sin^{2} (θ))

化简

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

2 + sin (θ) - 2 (1 - sin^{2} (θ))

乘开

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

- 2 (1 - sin^{2} (θ))

使用分配律:

a (b - c) = ab - a c

a = - 2, b = 1, c = sin^{2} (θ)

= - 2 \cdot 1 - (- 2) sin^{2} (θ)

使用加减运算法则

- (- a) = a

= - 2 \cdot 1 + 2 sin^{2} (θ)

数字相乘:

2 \cdot 1 = 2

= - 2 + 2 sin^{2} (θ)

= 2 + sin (θ) - 2 + 2 sin^{2} (θ)

化简

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

2 + sin (θ) - 2 + 2 sin^{2} (θ)

对同类项分组

= sin (θ) + 2 sin^{2} (θ) + 2 - 2

2 - 2 = 0

= 2 sin^{2} (θ) + sin (θ)

= 2 sin^{2} (θ) + sin (θ)

= 2 sin^{2} (θ) + sin (θ)

sin (θ) + 2 sin^{2} (θ) = 0

用替代法求解

sin (θ) + 2 sin^{2} (θ) = 0

令

: sin (θ) = u

u + 2 u^{2} = 0

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

u + 2 u^{2} = 0

改写成标准形式

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

2 u^{2} + u = 0

使用求根公式求解

2 u^{2} + u = 0

二次方程求根公式:

若

a = 2, b = 1, c = 0

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 2 \cdot 0}{2 \cdot 2}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 2 \cdot 0}{2 \cdot 2}

1^{2} - 4 \cdot 2 \cdot 0 = 1

1^{2} - 4 \cdot 2 \cdot 0

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 2 \cdot 0

使用法则

0 \cdot a = 0

= 1 - 0

数字相减:

1 - 0 = 1

= 1

使用法则

1 = 1

= 1

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

将解分隔开

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

u = \frac{- 1 + 1}{2 \cdot 2} : 0

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

数字相加/相减:

- 1 + 1 = 0

= \frac{0}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{0}{4}

使用法则

\frac{0}{a} = 0, a \neq = 0

= 0

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

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

数字相减:

- 1 - 1 = - 2

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

数字相乘:

2 \cdot 2 = 4

= \frac{- 2}{4}

使用分式法则:

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

= - \frac{2}{4}

约分:

2

= - \frac{1}{2}

二次方程组的解是:

u = 0, u = - \frac{1}{2}

u = sin (θ)

代回

sin (θ) = 0, sin (θ) = - \frac{1}{2}

sin (θ) = 0, sin (θ) = - \frac{1}{2}

sin (θ) = 0 : θ = 2 πn, θ = π + 2 πn

sin (θ) = 0

sin (θ) = 0

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 = 0 + 2 πn, θ = π + 2 πn

θ = 0 + 2 πn, θ = π + 2 πn

解

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn, θ = π + 2 πn

sin (θ) = - \frac{1}{2} : θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

sin (θ) = - \frac{1}{2}

sin (θ) = - \frac{1}{2}

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{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

合并所有解

θ = 2 πn, θ = π + 2 πn, θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

作图

流行的例子

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9-9sin(θ)=6cos^2(θ)

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