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受欢迎的三角函数 >

7sin^2(θ)-5sin(θ)=2

解答

7 sin^{2} (θ) - 5 sin (θ) = 2

解答

θ = \frac{π}{2} + 2 πn, θ = - 0.28975 \dots + 2 πn, θ = π + 0.28975 \dots + 2 πn

+1

度数

θ = 9 0^{\circ} + 36 0^{\circ} n, θ = - 16.60154 \dots^{\circ} + 36 0^{\circ} n, θ = 196.60154 \dots^{\circ} + 36 0^{\circ} n

求解步骤

7 sin^{2} (θ) - 5 sin (θ) = 2

用替代法求解

7 sin^{2} (θ) - 5 sin (θ) = 2

令

: sin (θ) = u

7 u^{2} - 5 u = 2

7 u^{2} - 5 u = 2 : u = 1, u = - \frac{2}{7}

7 u^{2} - 5 u = 2

将

2

para o lado esquerdo

7 u^{2} - 5 u = 2

两边减去

2

7 u^{2} - 5 u - 2 = 2 - 2

化简

7 u^{2} - 5 u - 2 = 0

7 u^{2} - 5 u - 2 = 0

使用求根公式求解

7 u^{2} - 5 u - 2 = 0

二次方程求根公式:

若

a = 7, b = - 5, c = - 2

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 7 ( - 2 )}{2 \cdot 7}

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 7 ( - 2 )}{2 \cdot 7}

(- 5)^{2} - 4 \cdot 7 (- 2) = 9

(- 5)^{2} - 4 \cdot 7 (- 2)

使用法则

- (- a) = a

= (- 5)^{2} + 4 \cdot 7 \cdot 2

使用指数法则:

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

若

n

是偶数

(- 5)^{2} = 5^{2}

= 5^{2} + 4 \cdot 7 \cdot 2

数字相乘:

4 \cdot 7 \cdot 2 = 56

= 5^{2} + 56

5^{2} = 25

= 25 + 56

数字相加:

25 + 56 = 81

= 81

因式分解数字:

81 = 9^{2}

= 9^{2}

使用根式运算法则:

n a^{n} = a

9^{2} = 9

= 9

u_{1, 2} = \frac{- ( - 5 ) \pm 9}{2 \cdot 7}

将解分隔开

u_{1} = \frac{- ( - 5 ) + 9}{2 \cdot 7}, u_{2} = \frac{- ( - 5 ) - 9}{2 \cdot 7}

u = \frac{- ( - 5 ) + 9}{2 \cdot 7} : 1

\frac{- ( - 5 ) + 9}{2 \cdot 7}

使用法则

- (- a) = a

= \frac{5 + 9}{2 \cdot 7}

数字相加:

5 + 9 = 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{- ( - 5 ) - 9}{2 \cdot 7} : - \frac{2}{7}

\frac{- ( - 5 ) - 9}{2 \cdot 7}

使用法则

- (- a) = a

= \frac{5 - 9}{2 \cdot 7}

数字相减:

5 - 9 = - 4

= \frac{- 4}{2 \cdot 7}

数字相乘:

2 \cdot 7 = 14

= \frac{- 4}{14}

使用分式法则:

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

= - \frac{4}{14}

约分:

2

= - \frac{2}{7}

二次方程组的解是:

u = 1, u = - \frac{2}{7}

u = sin (θ)

代回

sin (θ) = 1, sin (θ) = - \frac{2}{7}

sin (θ) = 1, sin (θ) = - \frac{2}{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{2}{7} : θ = arcsin (- \frac{2}{7}) + 2 πn, θ = π + arcsin (\frac{2}{7}) + 2 πn

sin (θ) = - \frac{2}{7}

使用反三角函数性质

sin (θ) = - \frac{2}{7}

sin (θ) = - \frac{2}{7}

的通解

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

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

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

合并所有解

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

以小数形式表示解

θ = \frac{π}{2} + 2 πn, θ = - 0.28975 \dots + 2 πn, θ = π + 0.28975 \dots + 2 πn

作图

流行的例子

sec(2x)=-(2/(sqrt(3)))

sec (2 x) = - (\frac{2}{3})

(e^{-ln(-(sin(θ))/(cos(θ)))})/2*sin(θ)=0

\frac{e ^{- l n (- \frac{s i n ( θ )}{c o s ( θ )})}}{2} \cdot sin (θ) = 0

7sin(2x)+12sin(x)=0

7 sin (2 x) + 12 sin (x) = 0

solvefor x,cos(x)=cos(2x)

so l v e f or x, cos (x) = cos (2 x)

cos(x)=0.50074242

cos (x) = 0.50074242