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受欢迎的三角函数 >

solvefor x,y^n+25y=cos(5x)

解答

求解

x, y^{n} + 25 y = cos (5 x)

解答

x = \frac{arccos ( y ^{n} + 25 y )}{5} + \frac{2 πk}{5}, x = - \frac{arccos ( y ^{n} + 25 y )}{5} + \frac{2 πk}{5}

求解步骤

y^{n} + 25 y = cos (5 x)

交换两边

cos (5 x) = y^{n} + 25 y

使用反三角函数性质

cos (5 x) = y^{n} + 25 y

cos (5 x) = y^{n} + 25 y

的通解

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

5 x = arccos (y^{n} + 25 y) + 2 πk, 5 x = - arccos (y^{n} + 25 y) + 2 πk

5 x = arccos (y^{n} + 25 y) + 2 πk, 5 x = - arccos (y^{n} + 25 y) + 2 πk

解

5 x = arccos (y^{n} + 25 y) + 2 πk : x = \frac{arccos ( y ^{n} + 25 y )}{5} + \frac{2 πk}{5}

5 x = arccos (y^{n} + 25 y) + 2 πk

两边除以

5

5 x = arccos (y^{n} + 25 y) + 2 πk

两边除以

5

\frac{5 x}{5} = \frac{arccos ( y ^{n} + 25 y )}{5} + \frac{2 πk}{5}

化简

x = \frac{arccos ( y ^{n} + 25 y )}{5} + \frac{2 πk}{5}

x = \frac{arccos ( y ^{n} + 25 y )}{5} + \frac{2 πk}{5}

解

5 x = - arccos (y^{n} + 25 y) + 2 πk : x = - \frac{arccos ( y ^{n} + 25 y )}{5} + \frac{2 πk}{5}

5 x = - arccos (y^{n} + 25 y) + 2 πk

两边除以

5

5 x = - arccos (y^{n} + 25 y) + 2 πk

两边除以

5

\frac{5 x}{5} = - \frac{arccos ( y ^{n} + 25 y )}{5} + \frac{2 πk}{5}

化简

x = - \frac{arccos ( y ^{n} + 25 y )}{5} + \frac{2 πk}{5}

x = - \frac{arccos ( y ^{n} + 25 y )}{5} + \frac{2 πk}{5}

x = \frac{arccos ( y ^{n} + 25 y )}{5} + \frac{2 πk}{5}, x = - \frac{arccos ( y ^{n} + 25 y )}{5} + \frac{2 πk}{5}

作图

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