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受欢迎的三角函数 >

solvefor x,7y^2+sin(3x)=12-y^4

解答

求解

x, 7 y^{2} + sin (3 x) = 12 - y^{4}

解答

x = \frac{arcsin ( 12 - y ^{4} - 7 y ^{2} )}{3} + \frac{2 πn}{3}, x = \frac{π}{3} + \frac{arcsin ( - 12 + y ^{4} + 7 y ^{2} )}{3} + \frac{2 πn}{3}

求解步骤

7 y^{2} + sin (3 x) = 12 - y^{4}

将

7 y^{2}

到右边

7 y^{2} + sin (3 x) = 12 - y^{4}

两边减去

7 y^{2}

7 y^{2} + sin (3 x) - 7 y^{2} = 12 - y^{4} - 7 y^{2}

化简

sin (3 x) = 12 - y^{4} - 7 y^{2}

sin (3 x) = 12 - y^{4} - 7 y^{2}

使用反三角函数性质

sin (3 x) = 12 - y^{4} - 7 y^{2}

sin (3 x) = 12 - y^{4} - 7 y^{2}

的通解

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π + arcsin (a) + 2 πn

3 x = arcsin (12 - y^{4} - 7 y^{2}) + 2 πn, 3 x = π + arcsin (- 12 + y^{4} + 7 y^{2}) + 2 πn

3 x = arcsin (12 - y^{4} - 7 y^{2}) + 2 πn, 3 x = π + arcsin (- 12 + y^{4} + 7 y^{2}) + 2 πn

解

3 x = arcsin (12 - y^{4} - 7 y^{2}) + 2 πn : x = \frac{arcsin ( 12 - y ^{4} - 7 y ^{2} )}{3} + \frac{2 πn}{3}

3 x = arcsin (12 - y^{4} - 7 y^{2}) + 2 πn

两边除以

3

3 x = arcsin (12 - y^{4} - 7 y^{2}) + 2 πn

两边除以

3

\frac{3 x}{3} = \frac{arcsin ( 12 - y ^{4} - 7 y ^{2} )}{3} + \frac{2 πn}{3}

化简

x = \frac{arcsin ( 12 - y ^{4} - 7 y ^{2} )}{3} + \frac{2 πn}{3}

x = \frac{arcsin ( 12 - y ^{4} - 7 y ^{2} )}{3} + \frac{2 πn}{3}

解

3 x = π + arcsin (- 12 + y^{4} + 7 y^{2}) + 2 πn : x = \frac{π}{3} + \frac{arcsin ( - 12 + y ^{4} + 7 y ^{2} )}{3} + \frac{2 πn}{3}

3 x = π + arcsin (- 12 + y^{4} + 7 y^{2}) + 2 πn

两边除以

3

3 x = π + arcsin (- 12 + y^{4} + 7 y^{2}) + 2 πn

两边除以

3

\frac{3 x}{3} = \frac{π}{3} + \frac{arcsin ( - 12 + y ^{4} + 7 y ^{2} )}{3} + \frac{2 πn}{3}

化简

x = \frac{π}{3} + \frac{arcsin ( - 12 + y ^{4} + 7 y ^{2} )}{3} + \frac{2 πn}{3}

x = \frac{π}{3} + \frac{arcsin ( - 12 + y ^{4} + 7 y ^{2} )}{3} + \frac{2 πn}{3}

x = \frac{arcsin ( 12 - y ^{4} - 7 y ^{2} )}{3} + \frac{2 πn}{3}, x = \frac{π}{3} + \frac{arcsin ( - 12 + y ^{4} + 7 y ^{2} )}{3} + \frac{2 πn}{3}

作图

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