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受欢迎的三角函数 >

7sin^2(θ)-3sin(θ)=4

解答

7 sin^{2} (θ) - 3 sin (θ) = 4

解答

θ = \frac{π}{2} + 2 πn, θ = - 0.60824 \dots + 2 πn, θ = π + 0.60824 \dots + 2 πn

+1

度数

θ = 9 0^{\circ} + 36 0^{\circ} n, θ = - 34.84990 \dots^{\circ} + 36 0^{\circ} n, θ = 214.84990 \dots^{\circ} + 36 0^{\circ} n

求解步骤

7 sin^{2} (θ) - 3 sin (θ) = 4

用替代法求解

7 sin^{2} (θ) - 3 sin (θ) = 4

令

: sin (θ) = u

7 u^{2} - 3 u = 4

7 u^{2} - 3 u = 4 : u = 1, u = - \frac{4}{7}

7 u^{2} - 3 u = 4

将

4

para o lado esquerdo

7 u^{2} - 3 u = 4

两边减去

4

7 u^{2} - 3 u - 4 = 4 - 4

化简

7 u^{2} - 3 u - 4 = 0

7 u^{2} - 3 u - 4 = 0

使用求根公式求解

7 u^{2} - 3 u - 4 = 0

二次方程求根公式:

若

a = 7, b = - 3, c = - 4

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 7 ( - 4 )}{2 \cdot 7}

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 7 ( - 4 )}{2 \cdot 7}

(- 3)^{2} - 4 \cdot 7 (- 4) = 11

(- 3)^{2} - 4 \cdot 7 (- 4)

使用法则

- (- a) = a

= (- 3)^{2} + 4 \cdot 7 \cdot 4

使用指数法则:

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

若

n

是偶数

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

= 3^{2} + 4 \cdot 7 \cdot 4

数字相乘:

4 \cdot 7 \cdot 4 = 112

= 3^{2} + 112

3^{2} = 9

= 9 + 112

数字相加:

9 + 112 = 121

= 121

因式分解数字:

121 = 1 1^{2}

= 1 1^{2}

使用根式运算法则:

n a^{n} = a

1 1^{2} = 11

= 11

u_{1, 2} = \frac{- ( - 3 ) \pm 11}{2 \cdot 7}

将解分隔开

u_{1} = \frac{- ( - 3 ) + 11}{2 \cdot 7}, u_{2} = \frac{- ( - 3 ) - 11}{2 \cdot 7}

u = \frac{- ( - 3 ) + 11}{2 \cdot 7} : 1

\frac{- ( - 3 ) + 11}{2 \cdot 7}

使用法则

- (- a) = a

= \frac{3 + 11}{2 \cdot 7}

数字相加:

3 + 11 = 14

= \frac{14}{2 \cdot 7}

数字相乘:

2 \cdot 7 = 14

= \frac{14}{14}

使用法则

\frac{a}{a} = 1

= 1

u = \frac{- ( - 3 ) - 11}{2 \cdot 7} : - \frac{4}{7}

\frac{- ( - 3 ) - 11}{2 \cdot 7}

使用法则

- (- a) = a

= \frac{3 - 11}{2 \cdot 7}

数字相减:

3 - 11 = - 8

= \frac{- 8}{2 \cdot 7}

数字相乘:

2 \cdot 7 = 14

= \frac{- 8}{14}

使用分式法则:

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

= - \frac{8}{14}

约分:

2

= - \frac{4}{7}

二次方程组的解是:

u = 1, u = - \frac{4}{7}

u = sin (θ)

代回

sin (θ) = 1, sin (θ) = - \frac{4}{7}

sin (θ) = 1, sin (θ) = - \frac{4}{7}

sin (θ) = 1 : θ = \frac{π}{2} + 2 πn

sin (θ) = 1

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{π}{2} + 2 πn

θ = \frac{π}{2} + 2 πn

sin (θ) = - \frac{4}{7} : θ = arcsin (- \frac{4}{7}) + 2 πn, θ = π + arcsin (\frac{4}{7}) + 2 πn

sin (θ) = - \frac{4}{7}

使用反三角函数性质

sin (θ) = - \frac{4}{7}

sin (θ) = - \frac{4}{7}

的通解

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

θ = arcsin (- \frac{4}{7}) + 2 πn, θ = π + arcsin (\frac{4}{7}) + 2 πn

θ = arcsin (- \frac{4}{7}) + 2 πn, θ = π + arcsin (\frac{4}{7}) + 2 πn

合并所有解

θ = \frac{π}{2} + 2 πn, θ = arcsin (- \frac{4}{7}) + 2 πn, θ = π + arcsin (\frac{4}{7}) + 2 πn

以小数形式表示解

θ = \frac{π}{2} + 2 πn, θ = - 0.60824 \dots + 2 πn, θ = π + 0.60824 \dots + 2 πn

作图

流行的例子

- 5 - 4 tan (θ) = - 1

sin (θ) = - \frac{7}{10}

10 cos (2 x) - 3 = 2

sin^2(θ)=9(cos(-θ)+1)

sin^{2} (θ) = 9 (cos (- θ) + 1)

sin(30x)=202.07

sin (30 x) = 202.07