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受欢迎的三角函数 >

3cos^2(θ)-1=0

解答

3 cos^{2} (θ) - 1 = 0

解答

θ = 0.95531 \dots + 2 πn, θ = 2 π - 0.95531 \dots + 2 πn, θ = 2.18627 \dots + 2 πn, θ = - 2.18627 \dots + 2 πn

+1

度数

θ = 54.73561 \dots^{\circ} + 36 0^{\circ} n, θ = 305.26438 \dots^{\circ} + 36 0^{\circ} n, θ = 125.26438 \dots^{\circ} + 36 0^{\circ} n, θ = - 125.26438 \dots^{\circ} + 36 0^{\circ} n

求解步骤

3 cos^{2} (θ) - 1 = 0

用替代法求解

3 cos^{2} (θ) - 1 = 0

令

: cos (θ) = u

3 u^{2} - 1 = 0

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

3 u^{2} - 1 = 0

将

1

到右边

3 u^{2} - 1 = 0

两边加上

1

3 u^{2} - 1 + 1 = 0 + 1

化简

3 u^{2} = 1

3 u^{2} = 1

两边除以

3

3 u^{2} = 1

两边除以

3

\frac{3 u ^{2}}{3} = \frac{1}{3}

化简

u^{2} = \frac{1}{3}

u^{2} = \frac{1}{3}

对于

x^{2} = f (a)

解为

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

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

u = cos (θ)

代回

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

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

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

cos (θ) = \frac{1}{3}

使用反三角函数性质

cos (θ) = \frac{1}{3}

cos (θ) = \frac{1}{3}

的通解

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

θ = arccos (\frac{1}{3}) + 2 πn, θ = 2 π - arccos (\frac{1}{3}) + 2 πn

θ = arccos (\frac{1}{3}) + 2 πn, θ = 2 π - arccos (\frac{1}{3}) + 2 πn

cos (θ) = - \frac{1}{3} : θ = arccos (- \frac{1}{3}) + 2 πn, θ = - arccos (- \frac{1}{3}) + 2 πn

cos (θ) = - \frac{1}{3}

使用反三角函数性质

cos (θ) = - \frac{1}{3}

cos (θ) = - \frac{1}{3}

的通解

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

θ = arccos (- \frac{1}{3}) + 2 πn, θ = - arccos (- \frac{1}{3}) + 2 πn

θ = arccos (- \frac{1}{3}) + 2 πn, θ = - arccos (- \frac{1}{3}) + 2 πn

合并所有解

θ = arccos (\frac{1}{3}) + 2 πn, θ = 2 π - arccos (\frac{1}{3}) + 2 πn, θ = arccos (- \frac{1}{3}) + 2 πn, θ = - arccos (- \frac{1}{3}) + 2 πn

以小数形式表示解

θ = 0.95531 \dots + 2 πn, θ = 2 π - 0.95531 \dots + 2 πn, θ = 2.18627 \dots + 2 πn, θ = - 2.18627 \dots + 2 πn

作图

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