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受欢迎的三角函数 >

16cos(2θ)-9=0

解答

16 cos (2 θ) - 9 = 0

解答

θ = \frac{0.97338 \dots}{2} + πn, θ = π - \frac{0.97338 \dots}{2} + πn

+1

度数

θ = 27.88556 \dots^{\circ} + 18 0^{\circ} n, θ = 152.11443 \dots^{\circ} + 18 0^{\circ} n

求解步骤

16 cos (2 θ) - 9 = 0

将

9

到右边

16 cos (2 θ) - 9 = 0

两边加上

9

16 cos (2 θ) - 9 + 9 = 0 + 9

化简

16 cos (2 θ) = 9

16 cos (2 θ) = 9

两边除以

16

16 cos (2 θ) = 9

两边除以

16

\frac{16 cos ( 2 θ )}{16} = \frac{9}{16}

化简

cos (2 θ) = \frac{9}{16}

cos (2 θ) = \frac{9}{16}

使用反三角函数性质

cos (2 θ) = \frac{9}{16}

cos (2 θ) = \frac{9}{16}

的通解

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

2 θ = arccos (\frac{9}{16}) + 2 πn, 2 θ = 2 π - arccos (\frac{9}{16}) + 2 πn

2 θ = arccos (\frac{9}{16}) + 2 πn, 2 θ = 2 π - arccos (\frac{9}{16}) + 2 πn

解

2 θ = arccos (\frac{9}{16}) + 2 πn : θ = \frac{arccos ( \frac{9}{16} )}{2} + πn

2 θ = arccos (\frac{9}{16}) + 2 πn

两边除以

2

2 θ = arccos (\frac{9}{16}) + 2 πn

两边除以

2

\frac{2 θ}{2} = \frac{arccos ( \frac{9}{16} )}{2} + \frac{2 πn}{2}

化简

θ = \frac{arccos ( \frac{9}{16} )}{2} + πn

θ = \frac{arccos ( \frac{9}{16} )}{2} + πn

解

2 θ = 2 π - arccos (\frac{9}{16}) + 2 πn : θ = π - \frac{arccos ( \frac{9}{16} )}{2} + πn

2 θ = 2 π - arccos (\frac{9}{16}) + 2 πn

两边除以

2

2 θ = 2 π - arccos (\frac{9}{16}) + 2 πn

两边除以

2

\frac{2 θ}{2} = \frac{2 π}{2} - \frac{arccos ( \frac{9}{16} )}{2} + \frac{2 πn}{2}

化简

θ = π - \frac{arccos ( \frac{9}{16} )}{2} + πn

θ = π - \frac{arccos ( \frac{9}{16} )}{2} + πn

θ = \frac{arccos ( \frac{9}{16} )}{2} + πn, θ = π - \frac{arccos ( \frac{9}{16} )}{2} + πn

以小数形式表示解

θ = \frac{0.97338 \dots}{2} + πn, θ = π - \frac{0.97338 \dots}{2} + πn

作图

流行的例子

(sin^2(θ)-2sin(θ)+1)/(sin(θ)-1)=-1

\frac{sin ^{2} ( θ ) - 2 sin ( θ ) + 1}{sin ( θ ) - 1} = - 1

solvefor w,100cos(10w)=50

so l v e f or w, 100 cos (10 w) = 50

cos (θ) = \frac{8}{25}

sin (2 x) = π

sin (x + 5 π) = 0