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受欢迎的三角函数 >

sin^2(θ)+sin(θ)-6=0

解答

sin^{2} (θ) + sin (θ) - 6 = 0

解答

θ \in R 无解

求解步骤

sin^{2} (θ) + sin (θ) - 6 = 0

用替代法求解

sin^{2} (θ) + sin (θ) - 6 = 0

令

: sin (θ) = u

u^{2} + u - 6 = 0

u^{2} + u - 6 = 0 : u = 2, u = - 3

u^{2} + u - 6 = 0

使用求根公式求解

u^{2} + u - 6 = 0

二次方程求根公式:

若

a = 1, b = 1, c = - 6

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 6 )}{2 \cdot 1}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 6 )}{2 \cdot 1}

1^{2} - 4 \cdot 1 \cdot (- 6) = 5

1^{2} - 4 \cdot 1 \cdot (- 6)

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 1 \cdot (- 6)

使用法则

- (- a) = a

= 1 + 4 \cdot 1 \cdot 6

数字相乘:

4 \cdot 1 \cdot 6 = 24

= 1 + 24

数字相加:

1 + 24 = 25

= 25

因式分解数字:

25 = 5^{2}

= 5^{2}

使用根式运算法则:

n a^{n} = a

5^{2} = 5

= 5

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

将解分隔开

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

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

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

数字相加/相减:

- 1 + 5 = 4

= \frac{4}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{4}{2}

数字相除:

\frac{4}{2} = 2

= 2

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

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

数字相减:

- 1 - 5 = - 6

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

数字相乘:

2 \cdot 1 = 2

= \frac{- 6}{2}

使用分式法则:

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

= - \frac{6}{2}

数字相除:

\frac{6}{2} = 3

= - 3

二次方程组的解是:

u = 2, u = - 3

u = sin (θ)

代回

sin (θ) = 2, sin (θ) = - 3

sin (θ) = 2, sin (θ) = - 3

sin (θ) = 2 :

无解

sin (θ) = 2

- 1 \leq sin (x) \leq 1

无解

sin (θ) = - 3 :

无解

sin (θ) = - 3

- 1 \leq sin (x) \leq 1

无解

合并所有解

θ \in R 无解

作图

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