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受欢迎的三角函数 >

sec(2θ)=csc(70)

解答

sec (2 θ) = csc (7 0^{\circ})

解答

θ = \frac{0.34906 \dots}{2} + 18 0^{\circ} n, θ = 18 0^{\circ} - \frac{0.34906 \dots}{2} + 18 0^{\circ} n

+1

弧度

θ = \frac{0.34906 \dots}{2} + πn, θ = π - \frac{0.34906 \dots}{2} + πn

求解步骤

sec (2 θ) = csc (7 0^{\circ})

使用反三角函数性质

sec (2 θ) = csc (7 0^{\circ})

sec (2 θ) = csc (7 0^{\circ})

的通解

sec (x) = a \Rightarrow x = arcsec (a) + 36 0^{\circ} n, x = 36 0^{\circ} - arcsec (a) + 36 0^{\circ} n

2 θ = arcsec (csc (7 0^{\circ})) + 36 0^{\circ} n, 2 θ = 36 0^{\circ} - arcsec (csc (7 0^{\circ})) + 36 0^{\circ} n

2 θ = arcsec (csc (7 0^{\circ})) + 36 0^{\circ} n, 2 θ = 36 0^{\circ} - arcsec (csc (7 0^{\circ})) + 36 0^{\circ} n

解

2 θ = arcsec (csc (7 0^{\circ})) + 36 0^{\circ} n : θ = \frac{arcsec ( csc ( 7 0 ^{\circ} ) )}{2} + 18 0^{\circ} n

2 θ = arcsec (csc (7 0^{\circ})) + 36 0^{\circ} n

两边除以

2

2 θ = arcsec (csc (7 0^{\circ})) + 36 0^{\circ} n

两边除以

2

\frac{2 θ}{2} = \frac{arcsec ( csc ( 7 0 ^{\circ} ) )}{2} + \frac{36 0 ^{\circ} n}{2}

化简

θ = \frac{arcsec ( csc ( 7 0 ^{\circ} ) )}{2} + 18 0^{\circ} n

θ = \frac{arcsec ( csc ( 7 0 ^{\circ} ) )}{2} + 18 0^{\circ} n

解

2 θ = 36 0^{\circ} - arcsec (csc (7 0^{\circ})) + 36 0^{\circ} n : θ = 18 0^{\circ} - \frac{arcsec ( csc ( 7 0 ^{\circ} ) )}{2} + 18 0^{\circ} n

2 θ = 36 0^{\circ} - arcsec (csc (7 0^{\circ})) + 36 0^{\circ} n

两边除以

2

2 θ = 36 0^{\circ} - arcsec (csc (7 0^{\circ})) + 36 0^{\circ} n

两边除以

2

\frac{2 θ}{2} = 18 0^{\circ} - \frac{arcsec ( csc ( 7 0 ^{\circ} ) )}{2} + \frac{36 0 ^{\circ} n}{2}

化简

θ = 18 0^{\circ} - \frac{arcsec ( csc ( 7 0 ^{\circ} ) )}{2} + 18 0^{\circ} n

θ = 18 0^{\circ} - \frac{arcsec ( csc ( 7 0 ^{\circ} ) )}{2} + 18 0^{\circ} n

θ = \frac{arcsec ( csc ( 7 0 ^{\circ} ) )}{2} + 18 0^{\circ} n, θ = 18 0^{\circ} - \frac{arcsec ( csc ( 7 0 ^{\circ} ) )}{2} + 18 0^{\circ} n

以小数形式表示解

θ = \frac{0.34906 \dots}{2} + 18 0^{\circ} n, θ = 18 0^{\circ} - \frac{0.34906 \dots}{2} + 18 0^{\circ} n

作图

流行的例子

sin(2x)-sin(4x)+sin(6x)=0

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