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受欢迎的三角函数 >

sin^2(θ)-2=sin(θ),-pi<= x<= 2pi

解答

sin^{2} (θ) - 2 = sin (θ), - π \leq x \leq 2 π

解答

θ = \frac{3 π}{2}, θ = - \frac{π}{2}

+1

度数

θ = 27 0^{\circ}, θ = - 9 0^{\circ}

求解步骤

sin^{2} (θ) - 2 = sin (θ), - π \leq x \leq 2 π

用替代法求解

sin^{2} (θ) - 2 = sin (θ)

令

: sin (θ) = u

u^{2} - 2 = u

u^{2} - 2 = u : u = 2, u = - 1

u^{2} - 2 = u

将

u

para o lado esquerdo

u^{2} - 2 = u

两边减去

u

u^{2} - 2 - u = u - u

化简

u^{2} - 2 - u = 0

u^{2} - 2 - u = 0

改写成标准形式

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

u^{2} - u - 2 = 0

使用求根公式求解

u^{2} - u - 2 = 0

二次方程求根公式:

若

a = 1, b = - 1, c = - 2

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

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

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

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

使用法则

- (- a) = a

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

(- 1)^{2} = 1

(- 1)^{2}

使用指数法则:

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

若

n

是偶数

(- 1)^{2} = 1^{2}

= 1^{2}

使用法则

1^{a} = 1

= 1

4 \cdot 1 \cdot 2 = 8

4 \cdot 1 \cdot 2

数字相乘:

4 \cdot 1 \cdot 2 = 8

= 8

= 1 + 8

数字相加:

1 + 8 = 9

= 9

因式分解数字:

9 = 3^{2}

= 3^{2}

使用根式运算法则:

n a^{n} = a

3^{2} = 3

= 3

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

将解分隔开

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

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

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

使用法则

- (- a) = a

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

数字相加:

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

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

使用法则

- (- a) = a

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

数字相减:

1 - 3 = - 2

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

数字相乘:

2 \cdot 1 = 2

= \frac{- 2}{2}

使用分式法则:

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

= - \frac{2}{2}

使用法则

\frac{a}{a} = 1

= - 1

二次方程组的解是:

u = 2, u = - 1

u = sin (θ)

代回

sin (θ) = 2, sin (θ) = - 1

sin (θ) = 2, sin (θ) = - 1

sin (θ) = 2, - π \leq x \leq 2 π :

无解

sin (θ) = 2, - π \leq x \leq 2 π

- 1 \leq sin (x) \leq 1

无解

sin (θ) = - 1, - π \leq x \leq 2 π : θ = \frac{3 π}{2}, θ = - \frac{π}{2}

sin (θ) = - 1, - π \leq x \leq 2 π

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

在

- π \leq x \leq 2 π

范围内的解

θ = \frac{3 π}{2}, θ = - \frac{π}{2}

合并所有解

θ = \frac{3 π}{2}, θ = - \frac{π}{2}

作图

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