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受欢迎的三角函数 >

solvefor x,ln(y)+y^2=sin(x)+c

解答

求解

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

解答

x = arcsin (ln (y) + y^{2} - c) + 2 πn, x = π + arcsin (- ln (y) - y^{2} + c) + 2 πn

求解步骤

ln (y) + y^{2} = sin (x) + c

交换两边

sin (x) + c = ln (y) + y^{2}

将

c

到右边

sin (x) + c = ln (y) + y^{2}

两边减去

c

sin (x) + c - c = ln (y) + y^{2} - c

化简

sin (x) = ln (y) + y^{2} - c

sin (x) = ln (y) + y^{2} - c

使用反三角函数性质

sin (x) = ln (y) + y^{2} - c

sin (x) = ln (y) + y^{2} - c

的通解

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

x = arcsin (ln (y) + y^{2} - c) + 2 πn, x = π + arcsin (- ln (y) - y^{2} + c) + 2 πn

x = arcsin (ln (y) + y^{2} - c) + 2 πn, x = π + arcsin (- ln (y) - y^{2} + c) + 2 πn

作图

流行的例子

5/2 =2+cos(x+(2pi)/3)

\frac{5}{2} = 2 + cos (x + \frac{2 π}{3})

sin(x)+cos(x)=sqrt((2+\sqrt{3))/2}

sin (x) + cos (x) = \frac{2 + 3}{2}

2 sec (x) = 0

sin^2(θ)cos^2(θ)=0,0<= θ<2pi

sin^{2} (θ) cos^{2} (θ) = 0, 0 \leq θ < 2 π

tan (θ) + 2 = 0