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受欢迎的三角函数 >

sin^2(θ)-3sin(θ)=0

解答

sin^{2} (θ) - 3 sin (θ) = 0

解答

θ = 2 πn, θ = π + 2 πn

+1

度数

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

求解步骤

sin^{2} (θ) - 3 sin (θ) = 0

用替代法求解

sin^{2} (θ) - 3 sin (θ) = 0

令

: sin (θ) = u

u^{2} - 3 u = 0

u^{2} - 3 u = 0 : u = 3, u = 0

u^{2} - 3 u = 0

使用求根公式求解

u^{2} - 3 u = 0

二次方程求根公式:

若

a = 1, b = - 3, c = 0

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

(- 3)^{2} - 4 \cdot 1 \cdot 0 = 3

(- 3)^{2} - 4 \cdot 1 \cdot 0

使用指数法则:

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

若

n

是偶数

(- 3)^{2} = 3^{2}

= 3^{2} - 4 \cdot 1 \cdot 0

使用法则

0 \cdot a = 0

= 3^{2} - 0

3^{2} - 0 = 3^{2}

= 3^{2}

使用根式运算法则:

假定

a \geq 0

= 3

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

将解分隔开

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

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

\frac{- ( - 3 ) + 3}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{3 + 3}{2 \cdot 1}

数字相加:

3 + 3 = 6

= \frac{6}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{6}{2}

数字相除:

\frac{6}{2} = 3

= 3

u = \frac{- ( - 3 ) - 3}{2 \cdot 1} : 0

\frac{- ( - 3 ) - 3}{2 \cdot 1}

使用法则

- (- a) = a

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

数字相减:

3 - 3 = 0

= \frac{0}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{0}{2}

使用法则

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

= 0

二次方程组的解是:

u = 3, u = 0

u = sin (θ)

代回

sin (θ) = 3, sin (θ) = 0

sin (θ) = 3, sin (θ) = 0

sin (θ) = 3 :

无解

sin (θ) = 3

- 1 \leq sin (x) \leq 1

无解

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

合并所有解

θ = 2 πn, θ = π + 2 πn

作图

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