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受欢迎的三角函数 >

-12cos^2(θ)+3=-3

解答

- 12 cos^{2} (θ) + 3 = - 3

解答

θ = 0.78539 \dots + 2 πn, θ = 2 π - 0.78539 \dots + 2 πn, θ = 2.35619 \dots + 2 πn, θ = - 2.35619 \dots + 2 πn

+1

度数

θ = 4 5^{\circ} + 36 0^{\circ} n, θ = 31 5^{\circ} + 36 0^{\circ} n, θ = 13 5^{\circ} + 36 0^{\circ} n, θ = - 13 5^{\circ} + 36 0^{\circ} n

求解步骤

- 12 cos^{2} (θ) + 3 = - 3

用替代法求解

- 12 cos^{2} (θ) + 3 = - 3

令

: cos (θ) = u

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

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

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

将

3

到右边

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

两边减去

3

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

化简

- 12 u^{2} = - 6

- 12 u^{2} = - 6

两边除以

- 12

- 12 u^{2} = - 6

两边除以

- 12

\frac{- 12 u ^{2}}{- 12} = \frac{- 6}{- 12}

化简

\frac{- 12 u ^{2}}{- 12} = \frac{- 6}{- 12}

化简

\frac{- 12 u ^{2}}{- 12} : u^{2}

\frac{- 12 u ^{2}}{- 12}

使用分式法则:

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

= \frac{12 u ^{2}}{12}

数字相除:

\frac{12}{12} = 1

= u^{2}

化简

\frac{- 6}{- 12} : \frac{1}{2}

\frac{- 6}{- 12}

使用分式法则:

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

= \frac{6}{12}

约分:

6

= \frac{1}{2}

u^{2} = \frac{1}{2}

u^{2} = \frac{1}{2}

u^{2} = \frac{1}{2}

对于

x^{2} = f (a)

解为

x = f (a), - f (a)

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

u = cos (θ)

代回

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

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

cos (θ) = \frac{1}{2} : θ = arccos (\frac{1}{2}) + 2 πn, θ = 2 π - arccos (\frac{1}{2}) + 2 πn

cos (θ) = \frac{1}{2}

使用反三角函数性质

cos (θ) = \frac{1}{2}

cos (θ) = \frac{1}{2}

的通解

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

θ = arccos (\frac{1}{2}) + 2 πn, θ = 2 π - arccos (\frac{1}{2}) + 2 πn

θ = arccos (\frac{1}{2}) + 2 πn, θ = 2 π - arccos (\frac{1}{2}) + 2 πn

cos (θ) = - \frac{1}{2} : θ = arccos (- \frac{1}{2}) + 2 πn, θ = - arccos (- \frac{1}{2}) + 2 πn

cos (θ) = - \frac{1}{2}

使用反三角函数性质

cos (θ) = - \frac{1}{2}

cos (θ) = - \frac{1}{2}

的通解

cos (x) = - a \Rightarrow x = arccos (- a) + 2 πn, x = - arccos (- a) + 2 πn

θ = arccos (- \frac{1}{2}) + 2 πn, θ = - arccos (- \frac{1}{2}) + 2 πn

θ = arccos (- \frac{1}{2}) + 2 πn, θ = - arccos (- \frac{1}{2}) + 2 πn

合并所有解

θ = arccos (\frac{1}{2}) + 2 πn, θ = 2 π - arccos (\frac{1}{2}) + 2 πn, θ = arccos (- \frac{1}{2}) + 2 πn, θ = - arccos (- \frac{1}{2}) + 2 πn

以小数形式表示解

θ = 0.78539 \dots + 2 πn, θ = 2 π - 0.78539 \dots + 2 πn, θ = 2.35619 \dots + 2 πn, θ = - 2.35619 \dots + 2 πn

作图

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