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受欢迎的三角函数 >

2sin^3(t)+sin^2(t)-2sin(t)-1=0

解答

2 sin^{3} (t) + sin^{2} (t) - 2 sin (t) - 1 = 0

解答

t = \frac{7 π}{6} + 2 πn, t = \frac{11 π}{6} + 2 πn, t = \frac{3 π}{2} + 2 πn, t = \frac{π}{2} + 2 πn

+1

度数

t = 21 0^{\circ} + 36 0^{\circ} n, t = 33 0^{\circ} + 36 0^{\circ} n, t = 27 0^{\circ} + 36 0^{\circ} n, t = 9 0^{\circ} + 36 0^{\circ} n

求解步骤

2 sin^{3} (t) + sin^{2} (t) - 2 sin (t) - 1 = 0

用替代法求解

2 sin^{3} (t) + sin^{2} (t) - 2 sin (t) - 1 = 0

令

: sin (t) = u

2 u^{3} + u^{2} - 2 u - 1 = 0

2 u^{3} + u^{2} - 2 u - 1 = 0 : u = - \frac{1}{2}, u = - 1, u = 1

2 u^{3} + u^{2} - 2 u - 1 = 0

因式分解

2 u^{3} + u^{2} - 2 u - 1 : (2 u + 1) (u + 1) (u - 1)

2 u^{3} + u^{2} - 2 u - 1

= (2 u^{3} + u^{2}) + (- 2 u - 1)

从

- 2 u - 1

分解出因式

- 1

:

- (2 u + 1)

- 2 u - 1

因式分解出通项

- 1

= - (2 u + 1)

从

2 u^{3} + u^{2}

分解出因式

u^{2}

:

u^{2} (2 u + 1)

2 u^{3} + u^{2}

使用指数法则:

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

u^{3} = u u^{2}

= 2 u u^{2} + u^{2}

因式分解出通项

u^{2}

= u^{2} (2 u + 1)

= - (2 u + 1) + u^{2} (2 u + 1)

因式分解出通项

2 u + 1

= (2 u + 1) (u^{2} - 1)

分解

u^{2} - 1 : (u + 1) (u - 1)

u^{2} - 1

将

1

改写为

1^{2}

= u^{2} - 1^{2}

使用平方差公式:

x^{2} - y^{2} = (x + y) (x - y)

u^{2} - 1^{2} = (u + 1) (u - 1)

= (u + 1) (u - 1)

= (2 u + 1) (u + 1) (u - 1)

(2 u + 1) (u + 1) (u - 1) = 0

使用零因数法则： If

ab = 0

then

a = 0

or

b = 0

2 u + 1 = 0 or u + 1 = 0 or u - 1 = 0

解

2 u + 1 = 0 : u = - \frac{1}{2}

2 u + 1 = 0

将

1

到右边

2 u + 1 = 0

两边减去

1

2 u + 1 - 1 = 0 - 1

化简

2 u = - 1

2 u = - 1

两边除以

2

2 u = - 1

两边除以

2

\frac{2 u}{2} = \frac{- 1}{2}

化简

u = - \frac{1}{2}

u = - \frac{1}{2}

解

u + 1 = 0 : u = - 1

u + 1 = 0

将

1

到右边

u + 1 = 0

两边减去

1

u + 1 - 1 = 0 - 1

化简

u = - 1

u = - 1

解

u - 1 = 0 : u = 1

u - 1 = 0

将

1

到右边

u - 1 = 0

两边加上

1

u - 1 + 1 = 0 + 1

化简

u = 1

u = 1

解为

u = - \frac{1}{2}, u = - 1, u = 1

u = sin (t)

代回

sin (t) = - \frac{1}{2}, sin (t) = - 1, sin (t) = 1

sin (t) = - \frac{1}{2}, sin (t) = - 1, sin (t) = 1

sin (t) = - \frac{1}{2} : t = \frac{7 π}{6} + 2 πn, t = \frac{11 π}{6} + 2 πn

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

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

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}

t = \frac{7 π}{6} + 2 πn, t = \frac{11 π}{6} + 2 πn

t = \frac{7 π}{6} + 2 πn, t = \frac{11 π}{6} + 2 πn

sin (t) = - 1 : t = \frac{3 π}{2} + 2 πn

sin (t) = - 1

sin (t) = - 1

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}

t = \frac{3 π}{2} + 2 πn

t = \frac{3 π}{2} + 2 πn

sin (t) = 1 : t = \frac{π}{2} + 2 πn

sin (t) = 1

sin (t) = 1

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}

t = \frac{π}{2} + 2 πn

t = \frac{π}{2} + 2 πn

合并所有解

t = \frac{7 π}{6} + 2 πn, t = \frac{11 π}{6} + 2 πn, t = \frac{3 π}{2} + 2 πn, t = \frac{π}{2} + 2 πn

作图

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