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受欢迎的三角函数 >

cos^2(x)=((4k-5))/(10)

解答

cos^{2} (x) = \frac{( 4 k - 5 )}{10}

解答

x = arccos (\frac{4 k - 5}{10}) + 2 πn, x = - arccos (\frac{4 k - 5}{10}) + 2 πn, x = arccos (- \frac{4 k - 5}{10}) + 2 πn, x = - arccos (- \frac{4 k - 5}{10}) + 2 πn

求解步骤

cos^{2} (x) = \frac{( 4 k - 5 )}{10}

用替代法求解

cos^{2} (x) = \frac{4 k - 5}{10}

令

: cos (x) = u

u^{2} = \frac{4 k - 5}{10}

u^{2} = \frac{4 k - 5}{10} : u = \frac{4 k - 5}{10}, u = - \frac{4 k - 5}{10}

u^{2} = \frac{4 k - 5}{10}

对于

x^{2} = f (a)

解为

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

u = \frac{4 k - 5}{10}, u = - \frac{4 k - 5}{10}

u = cos (x)

代回

cos (x) = \frac{4 k - 5}{10}, cos (x) = - \frac{4 k - 5}{10}

cos (x) = \frac{4 k - 5}{10}, cos (x) = - \frac{4 k - 5}{10}

cos (x) = \frac{4 k - 5}{10} : x = arccos (\frac{4 k - 5}{10}) + 2 πn, x = - arccos (\frac{4 k - 5}{10}) + 2 πn

cos (x) = \frac{4 k - 5}{10}

使用反三角函数性质

cos (x) = \frac{4 k - 5}{10}

cos (x) = \frac{4 k - 5}{10}

的通解

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

x = arccos (\frac{4 k - 5}{10}) + 2 πn, x = - arccos (\frac{4 k - 5}{10}) + 2 πn

x = arccos (\frac{4 k - 5}{10}) + 2 πn, x = - arccos (\frac{4 k - 5}{10}) + 2 πn

cos (x) = - \frac{4 k - 5}{10} : x = arccos (- \frac{4 k - 5}{10}) + 2 πn, x = - arccos (- \frac{4 k - 5}{10}) + 2 πn

cos (x) = - \frac{4 k - 5}{10}

使用反三角函数性质

cos (x) = - \frac{4 k - 5}{10}

cos (x) = - \frac{4 k - 5}{10}

的通解

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

x = arccos (- \frac{4 k - 5}{10}) + 2 πn, x = - arccos (- \frac{4 k - 5}{10}) + 2 πn

x = arccos (- \frac{4 k - 5}{10}) + 2 πn, x = - arccos (- \frac{4 k - 5}{10}) + 2 πn

合并所有解

x = arccos (\frac{4 k - 5}{10}) + 2 πn, x = - arccos (\frac{4 k - 5}{10}) + 2 πn, x = arccos (- \frac{4 k - 5}{10}) + 2 πn, x = - arccos (- \frac{4 k - 5}{10}) + 2 πn

作图

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