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受欢迎的三角函数 >

solvefor x,y^2(2+sin(x))=1

解答

求解

x, y^{2} (2 + sin (x)) = 1

解答

x = arcsin (\frac{1}{y ^{2}} - 2) + 2 πn, x = π + arcsin (- \frac{1}{y ^{2}} + 2) + 2 πn

求解步骤

y^{2} (2 + sin (x)) = 1

两边除以

y^{2}; y \neq = 0

y^{2} (2 + sin (x)) = 1

两边除以

y^{2}; y \neq = 0

\frac{y ^{2} ( 2 + sin ( x ) )}{y ^{2}} = \frac{1}{y ^{2}}; y \neq = 0

化简

2 + sin (x) = \frac{1}{y ^{2}}; y \neq = 0

2 + sin (x) = \frac{1}{y ^{2}}; y \neq = 0

将

2

到右边

2 + sin (x) = \frac{1}{y ^{2}}; y \neq = 0

两边减去

2

2 + sin (x) - 2 = \frac{1}{y ^{2}} - 2; y \neq = 0

化简

sin (x) = \frac{1}{y ^{2}} - 2; y \neq = 0

sin (x) = \frac{1}{y ^{2}} - 2; y \neq = 0

使用反三角函数性质

sin (x) = \frac{1}{y ^{2}} - 2

sin (x) = \frac{1}{y ^{2}} - 2

的通解

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

x = arcsin (\frac{1}{y ^{2}} - 2) + 2 πn, x = π + arcsin (- \frac{1}{y ^{2}} + 2) + 2 πn

x = arcsin (\frac{1}{y ^{2}} - 2) + 2 πn, x = π + arcsin (- \frac{1}{y ^{2}} + 2) + 2 πn

作图

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