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受欢迎的三角函数 >

1.16cos(θ)+0.532cos(θ)-0.557=0

解答

1.16 cos (θ) + 0.532 cos (θ) - 0.557 = 0

解答

θ = 1.23534 \dots + 2 πn, θ = 2 π - 1.23534 \dots + 2 πn

+1

度数

θ = 70.78000 \dots^{\circ} + 36 0^{\circ} n, θ = 289.21999 \dots^{\circ} + 36 0^{\circ} n

求解步骤

1.16 cos (θ) + 0.532 cos (θ) - 0.557 = 0

化简

1.16 cos (θ) + 0.532 cos (θ) - 0.557 : 1.692 cos (θ) - 0.557

1.16 cos (θ) + 0.532 cos (θ) - 0.557

同类项相加:

1.16 cos (θ) + 0.532 cos (θ) = 1.692 cos (θ)

= 1.692 cos (θ) - 0.557

1.692 cos (θ) - 0.557 = 0

在两边乘以

1000

1.692 cos (θ) - 0.557 = 0

To eliminate decimal points, multiply by 10 for every digit after the decimal pointThere are

3

digits to the right of the decimal point, therefore multiply by

1000

1.692 cos (θ) \cdot 1000 - 0.557 \cdot 1000 = 0 \cdot 1000

整理后得

1692 cos (θ) - 557 = 0

1692 cos (θ) - 557 = 0

将

557

到右边

1692 cos (θ) - 557 = 0

两边加上

557

1692 cos (θ) - 557 + 557 = 0 + 557

化简

1692 cos (θ) = 557

1692 cos (θ) = 557

两边除以

1692

1692 cos (θ) = 557

两边除以

1692

\frac{1692 cos ( θ )}{1692} = \frac{557}{1692}

化简

cos (θ) = \frac{557}{1692}

cos (θ) = \frac{557}{1692}

使用反三角函数性质

cos (θ) = \frac{557}{1692}

cos (θ) = \frac{557}{1692}

的通解

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

θ = arccos (\frac{557}{1692}) + 2 πn, θ = 2 π - arccos (\frac{557}{1692}) + 2 πn

θ = arccos (\frac{557}{1692}) + 2 πn, θ = 2 π - arccos (\frac{557}{1692}) + 2 πn

以小数形式表示解

θ = 1.23534 \dots + 2 πn, θ = 2 π - 1.23534 \dots + 2 πn

作图

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