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受欢迎的三角函数 >

5sin^2(θ)+2sin(θ)=0

解答

5 sin^{2} (θ) + 2 sin (θ) = 0

解答

θ = 2 πn, θ = π + 2 πn, θ = - 0.41151 \dots + 2 πn, θ = π + 0.41151 \dots + 2 πn

+1

度数

θ = 0^{\circ} + 36 0^{\circ} n, θ = 18 0^{\circ} + 36 0^{\circ} n, θ = - 23.57817 \dots^{\circ} + 36 0^{\circ} n, θ = 203.57817 \dots^{\circ} + 36 0^{\circ} n

求解步骤

5 sin^{2} (θ) + 2 sin (θ) = 0

用替代法求解

5 sin^{2} (θ) + 2 sin (θ) = 0

令

: sin (θ) = u

5 u^{2} + 2 u = 0

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

5 u^{2} + 2 u = 0

使用求根公式求解

5 u^{2} + 2 u = 0

二次方程求根公式:

若

a = 5, b = 2, c = 0

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 5 \cdot 0}{2 \cdot 5}

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 5 \cdot 0}{2 \cdot 5}

2^{2} - 4 \cdot 5 \cdot 0 = 2

2^{2} - 4 \cdot 5 \cdot 0

使用法则

0 \cdot a = 0

= 2^{2} - 0

2^{2} - 0 = 2^{2}

= 2^{2}

使用根式运算法则:

n a^{n} = a,

假定

a \geq 0

= 2

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

将解分隔开

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

u = \frac{- 2 + 2}{2 \cdot 5} : 0

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

数字相加/相减:

- 2 + 2 = 0

= \frac{0}{2 \cdot 5}

数字相乘:

2 \cdot 5 = 10

= \frac{0}{10}

使用法则

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

= 0

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

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

数字相减:

- 2 - 2 = - 4

= \frac{- 4}{2 \cdot 5}

数字相乘:

2 \cdot 5 = 10

= \frac{- 4}{10}

使用分式法则:

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

= - \frac{4}{10}

约分:

2

= - \frac{2}{5}

二次方程组的解是:

u = 0, u = - \frac{2}{5}

u = sin (θ)

代回

sin (θ) = 0, sin (θ) = - \frac{2}{5}

sin (θ) = 0, sin (θ) = - \frac{2}{5}

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

sin (θ) = - \frac{2}{5}

使用反三角函数性质

sin (θ) = - \frac{2}{5}

sin (θ) = - \frac{2}{5}

的通解

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

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

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

合并所有解

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

以小数形式表示解

θ = 2 πn, θ = π + 2 πn, θ = - 0.41151 \dots + 2 πn, θ = π + 0.41151 \dots + 2 πn

作图

流行的例子

cos(θ)cos(3θ)-sin(θ)sin(3θ)=0

cos (θ) cos (3 θ) - sin (θ) sin (3 θ) = 0

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1 - 2 sin (θ) = cos (2 θ), 0 \leq θ \leq 2 π

2sin^2(θ)+5sin(θ)-12=0

2 sin^{2} (θ) + 5 sin (θ) - 12 = 0

cos(2θ)= 3/5 ,90<= θ<= 180

cos (2 θ) = \frac{3}{5}, 9 0^{\circ} \leq θ \leq 18 0^{\circ}

sin^2(x)-4cos(x)-4=0,0<= x<= 2pi

sin^{2} (x) - 4 cos (x) - 4 = 0, 0 \leq x \leq 2 π