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受欢迎的三角函数 >

sin^2(x)cos(x)+sin(x)=0

解答

sin^{2} (x) cos (x) + sin (x) = 0

解答

x = 2 πn, x = π + 2 πn

+1

度数

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

求解步骤

sin^{2} (x) cos (x) + sin (x) = 0

分解

sin^{2} (x) cos (x) + sin (x) : sin (x) (sin (x) cos (x) + 1)

sin^{2} (x) cos (x) + sin (x)

使用指数法则:

a^{b + c} = a^{b} a^{c}

sin^{2} (x) cos (x) = sin (x) sin (x)

= sin (x) sin (x) + sin (x)

因式分解出通项

sin (x)

= sin (x) (sin (x) cos (x) + 1)

sin (x) (sin (x) cos (x) + 1) = 0

分别求解每个部分

sin (x) = 0 or sin (x) cos (x) + 1 = 0

sin (x) = 0 : x = 2 πn, x = π + 2 πn

sin (x) = 0

sin (x) = 0

的通解

sin (x)

周期表（周期为

2 πn

"）：

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

解

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

sin (x) cos (x) + 1 = 0 :

无解

sin (x) cos (x) + 1 = 0

使用三角恒等式改写

sin (x) cos (x) + 1

使用倍角公式:

2 sin (x) cos (x) = sin (2 x)

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

= 1 + \frac{sin ( 2 x )}{2}

1 + \frac{sin ( 2 x )}{2} = 0

将

1

到右边

1 + \frac{sin ( 2 x )}{2} = 0

两边减去

1

1 + \frac{sin ( 2 x )}{2} - 1 = 0 - 1

化简

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

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

在两边乘以

2

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

在两边乘以

2

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

化简

sin (2 x) = - 2

sin (2 x) = - 2

- 1 \leq sin (x) \leq 1

无解

合并所有解

x = 2 πn, x = π + 2 πn

作图

流行的例子

solvefor x,8sec(x)+8tan(x)=8

so l v e f or x, 8 sec (x) + 8 tan (x) = 8

3sin(2x)=5cos^2(2x)

3 sin (2 x) = 5 cos^{2} (2 x)

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

so l v e f or x, y^{n} + 25 y = cos (5 x)

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

sin^2(x)+sin(x)=3sin^2(x)

sin^{2} (x) + sin (x) = 3 sin^{2} (x)