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受欢迎的三角函数 >

9cos(2θ)=8

解答

9 cos (2 θ) = 8

解答

θ = \frac{0.47588 \dots}{2} + πn, θ = π - \frac{0.47588 \dots}{2} + πn

+1

度数

θ = 13.63302 \dots^{\circ} + 18 0^{\circ} n, θ = 166.36697 \dots^{\circ} + 18 0^{\circ} n

求解步骤

9 cos (2 θ) = 8

两边除以

9

9 cos (2 θ) = 8

两边除以

9

\frac{9 cos ( 2 θ )}{9} = \frac{8}{9}

化简

cos (2 θ) = \frac{8}{9}

cos (2 θ) = \frac{8}{9}

使用反三角函数性质

cos (2 θ) = \frac{8}{9}

cos (2 θ) = \frac{8}{9}

的通解

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

2 θ = arccos (\frac{8}{9}) + 2 πn, 2 θ = 2 π - arccos (\frac{8}{9}) + 2 πn

2 θ = arccos (\frac{8}{9}) + 2 πn, 2 θ = 2 π - arccos (\frac{8}{9}) + 2 πn

解

2 θ = arccos (\frac{8}{9}) + 2 πn : θ = \frac{arccos ( \frac{8}{9} )}{2} + πn

2 θ = arccos (\frac{8}{9}) + 2 πn

两边除以

2

2 θ = arccos (\frac{8}{9}) + 2 πn

两边除以

2

\frac{2 θ}{2} = \frac{arccos ( \frac{8}{9} )}{2} + \frac{2 πn}{2}

化简

θ = \frac{arccos ( \frac{8}{9} )}{2} + πn

θ = \frac{arccos ( \frac{8}{9} )}{2} + πn

解

2 θ = 2 π - arccos (\frac{8}{9}) + 2 πn : θ = π - \frac{arccos ( \frac{8}{9} )}{2} + πn

2 θ = 2 π - arccos (\frac{8}{9}) + 2 πn

两边除以

2

2 θ = 2 π - arccos (\frac{8}{9}) + 2 πn

两边除以

2

\frac{2 θ}{2} = \frac{2 π}{2} - \frac{arccos ( \frac{8}{9} )}{2} + \frac{2 πn}{2}

化简

θ = π - \frac{arccos ( \frac{8}{9} )}{2} + πn

θ = π - \frac{arccos ( \frac{8}{9} )}{2} + πn

θ = \frac{arccos ( \frac{8}{9} )}{2} + πn, θ = π - \frac{arccos ( \frac{8}{9} )}{2} + πn

以小数形式表示解

θ = \frac{0.47588 \dots}{2} + πn, θ = π - \frac{0.47588 \dots}{2} + πn

作图

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