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受欢迎的三角函数 >

3cos^2(θ)-sin(θ)=1

解答

3 cos^{2} (θ) - sin (θ) = 1

解答

θ = \frac{3 π}{2} + 2 πn, θ = 0.72972 \dots + 2 πn, θ = π - 0.72972 \dots + 2 πn

+1

度数

θ = 27 0^{\circ} + 36 0^{\circ} n, θ = 41.81031 \dots^{\circ} + 36 0^{\circ} n, θ = 138.18968 \dots^{\circ} + 36 0^{\circ} n

求解步骤

3 cos^{2} (θ) - sin (θ) = 1

两边减去

1

3 cos^{2} (θ) - sin (θ) - 1 = 0

使用三角恒等式改写

- 1 - sin (θ) + 3 cos^{2} (θ)

使用毕达哥拉斯恒等式:

cos^{2} (x) + sin^{2} (x) = 1

cos^{2} (x) = 1 - sin^{2} (x)

= - 1 - sin (θ) + 3 (1 - sin^{2} (θ))

化简

- 1 - sin (θ) + 3 (1 - sin^{2} (θ)) : - 3 sin^{2} (θ) - sin (θ) + 2

- 1 - sin (θ) + 3 (1 - sin^{2} (θ))

乘开

3 (1 - sin^{2} (θ)) : 3 - 3 sin^{2} (θ)

3 (1 - sin^{2} (θ))

使用分配律:

a (b - c) = ab - a c

a = 3, b = 1, c = sin^{2} (θ)

= 3 \cdot 1 - 3 sin^{2} (θ)

数字相乘:

3 \cdot 1 = 3

= 3 - 3 sin^{2} (θ)

= - 1 - sin (θ) + 3 - 3 sin^{2} (θ)

化简

- 1 - sin (θ) + 3 - 3 sin^{2} (θ) : - 3 sin^{2} (θ) - sin (θ) + 2

- 1 - sin (θ) + 3 - 3 sin^{2} (θ)

对同类项分组

= - sin (θ) - 3 sin^{2} (θ) - 1 + 3

数字相加/相减:

- 1 + 3 = 2

= - 3 sin^{2} (θ) - sin (θ) + 2

= - 3 sin^{2} (θ) - sin (θ) + 2

= - 3 sin^{2} (θ) - sin (θ) + 2

2 - sin (θ) - 3 sin^{2} (θ) = 0

用替代法求解

2 - sin (θ) - 3 sin^{2} (θ) = 0

令

: sin (θ) = u

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

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

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

改写成标准形式

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

- 3 u^{2} - u + 2 = 0

使用求根公式求解

- 3 u^{2} - u + 2 = 0

二次方程求根公式:

若

a = - 3, b = - 1, c = 2

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

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

(- 1)^{2} - 4 (- 3) \cdot 2 = 5

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

使用法则

- (- a) = a

= (- 1)^{2} + 4 \cdot 3 \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 3 \cdot 2 = 24

4 \cdot 3 \cdot 2

数字相乘:

4 \cdot 3 \cdot 2 = 24

= 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 ( - 3 )}

将解分隔开

u_{1} = \frac{- ( - 1 ) + 5}{2 ( - 3 )}, u_{2} = \frac{- ( - 1 ) - 5}{2 ( - 3 )}

u = \frac{- ( - 1 ) + 5}{2 ( - 3 )} : - 1

\frac{- ( - 1 ) + 5}{2 ( - 3 )}

去除括号:

(- a) = - a, - (- a) = a

= \frac{1 + 5}{- 2 \cdot 3}

数字相加:

1 + 5 = 6

= \frac{6}{- 2 \cdot 3}

数字相乘:

2 \cdot 3 = 6

= \frac{6}{- 6}

使用分式法则:

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

= - \frac{6}{6}

使用法则

\frac{a}{a} = 1

= - 1

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

\frac{- ( - 1 ) - 5}{2 ( - 3 )}

去除括号:

(- a) = - a, - (- a) = a

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

数字相减:

1 - 5 = - 4

= \frac{- 4}{- 2 \cdot 3}

数字相乘:

2 \cdot 3 = 6

= \frac{- 4}{- 6}

使用分式法则:

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

= \frac{4}{6}

约分:

2

= \frac{2}{3}

二次方程组的解是:

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

u = sin (θ)

代回

sin (θ) = - 1, sin (θ) = \frac{2}{3}

sin (θ) = - 1, sin (θ) = \frac{2}{3}

sin (θ) = - 1 : θ = \frac{3 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{3 π}{2} + 2 πn

θ = \frac{3 π}{2} + 2 πn

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

sin (θ) = \frac{2}{3}

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

合并所有解

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

以小数形式表示解

θ = \frac{3 π}{2} + 2 πn, θ = 0.72972 \dots + 2 πn, θ = π - 0.72972 \dots + 2 πn

作图

流行的例子

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