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受欢迎的三角函数 >

6sin^2(θ)-sin(θ)-2=0,0<= θ<= 2pi

解答

6 sin^{2} (θ) - sin (θ) - 2 = 0, 0 \leq θ \leq 2 π

解答

θ = 0.72972 \dots, θ = π - 0.72972 \dots, θ = \frac{7 π}{6}, θ = \frac{11 π}{6}

+1

度数

θ = 41.81031 \dots^{\circ}, θ = 138.18968 \dots^{\circ}, θ = 21 0^{\circ}, θ = 33 0^{\circ}

求解步骤

6 sin^{2} (θ) - sin (θ) - 2 = 0, 0 \leq θ \leq 2 π

用替代法求解

6 sin^{2} (θ) - sin (θ) - 2 = 0

令

: sin (θ) = u

6 u^{2} - u - 2 = 0

6 u^{2} - u - 2 = 0 : u = \frac{2}{3}, u = - \frac{1}{2}

6 u^{2} - u - 2 = 0

使用求根公式求解

6 u^{2} - u - 2 = 0

二次方程求根公式:

若

a = 6, b = - 1, c = - 2

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

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

(- 1)^{2} - 4 \cdot 6 (- 2) = 7

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

使用法则

- (- a) = a

= (- 1)^{2} + 4 \cdot 6 \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 6 \cdot 2 = 48

4 \cdot 6 \cdot 2

数字相乘:

4 \cdot 6 \cdot 2 = 48

= 48

= 1 + 48

数字相加:

1 + 48 = 49

= 49

因式分解数字:

49 = 7^{2}

= 7^{2}

使用根式运算法则:

n a^{n} = a

7^{2} = 7

= 7

u_{1, 2} = \frac{- ( - 1 ) \pm 7}{2 \cdot 6}

将解分隔开

u_{1} = \frac{- ( - 1 ) + 7}{2 \cdot 6}, u_{2} = \frac{- ( - 1 ) - 7}{2 \cdot 6}

u = \frac{- ( - 1 ) + 7}{2 \cdot 6} : \frac{2}{3}

\frac{- ( - 1 ) + 7}{2 \cdot 6}

使用法则

- (- a) = a

= \frac{1 + 7}{2 \cdot 6}

数字相加:

1 + 7 = 8

= \frac{8}{2 \cdot 6}

数字相乘:

2 \cdot 6 = 12

= \frac{8}{12}

约分:

4

= \frac{2}{3}

u = \frac{- ( - 1 ) - 7}{2 \cdot 6} : - \frac{1}{2}

\frac{- ( - 1 ) - 7}{2 \cdot 6}

使用法则

- (- a) = a

= \frac{1 - 7}{2 \cdot 6}

数字相减:

1 - 7 = - 6

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

数字相乘:

2 \cdot 6 = 12

= \frac{- 6}{12}

使用分式法则:

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

= - \frac{6}{12}

约分:

6

= - \frac{1}{2}

二次方程组的解是:

u = \frac{2}{3}, u = - \frac{1}{2}

u = sin (θ)

代回

sin (θ) = \frac{2}{3}, sin (θ) = - \frac{1}{2}

sin (θ) = \frac{2}{3}, sin (θ) = - \frac{1}{2}

sin (θ) = \frac{2}{3}, 0 \leq θ \leq 2 π : θ = arcsin (\frac{2}{3}), θ = π - arcsin (\frac{2}{3})

sin (θ) = \frac{2}{3}, 0 \leq θ \leq 2 π

使用反三角函数性质

sin (θ) = \frac{2}{3}

sin (θ) = \frac{2}{3}

的通解

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

θ = arcsin (\frac{2}{3}) + 2 πn, θ = π - arcsin (\frac{2}{3}) + 2 πn

θ = arcsin (\frac{2}{3}) + 2 πn, θ = π - arcsin (\frac{2}{3}) + 2 πn

在

0 \leq θ \leq 2 π

范围内的解

θ = arcsin (\frac{2}{3}), θ = π - arcsin (\frac{2}{3})

sin (θ) = - \frac{1}{2}, 0 \leq θ \leq 2 π : θ = \frac{7 π}{6}, θ = \frac{11 π}{6}

sin (θ) = - \frac{1}{2}, 0 \leq θ \leq 2 π

sin (θ) = - \frac{1}{2}

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{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

在

0 \leq θ \leq 2 π

范围内的解

θ = \frac{7 π}{6}, θ = \frac{11 π}{6}

合并所有解

θ = arcsin (\frac{2}{3}), θ = π - arcsin (\frac{2}{3}), θ = \frac{7 π}{6}, θ = \frac{11 π}{6}

以小数形式表示解

θ = 0.72972 \dots, θ = π - 0.72972 \dots, θ = \frac{7 π}{6}, θ = \frac{11 π}{6}

作图

流行的例子

3sqrt(3)cot(x)=3

33 cot (x) = 3

2 tan^{2} (x) - 6 = 0

0 = 2 sin (x) - 1

solvefor t,x=cos(2t)

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

-9sin^2(θ)-3cos(θ)+2=-5

- 9 sin^{2} (θ) - 3 cos (θ) + 2 = - 5