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受欢迎的三角函数 >

2sin(x)cos^2(x)-1+2cos^2(x)-sin(x)=0

解答

2 sin (x) cos^{2} (x) - 1 + 2 cos^{2} (x) - sin (x) = 0

解答

x = \frac{3 π}{2} + 2 πn, x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn, x = \frac{π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn

+1

度数

x = 27 0^{\circ} + 36 0^{\circ} n, x = 22 5^{\circ} + 36 0^{\circ} n, x = 31 5^{\circ} + 36 0^{\circ} n, x = 4 5^{\circ} + 36 0^{\circ} n, x = 13 5^{\circ} + 36 0^{\circ} n

求解步骤

2 sin (x) cos^{2} (x) - 1 + 2 cos^{2} (x) - sin (x) = 0

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

cos^{2} (x) = 1 - sin^{2} (x)

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

化简

- 1 - sin (x) + 2 (1 - sin^{2} (x)) + 2 (1 - sin^{2} (x)) sin (x) : sin (x) - 2 sin^{3} (x) - 2 sin^{2} (x) + 1

- 1 - sin (x) + 2 (1 - sin^{2} (x)) + 2 (1 - sin^{2} (x)) sin (x)

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

乘开

2 (1 - sin^{2} (x)) : 2 - 2 sin^{2} (x)

2 (1 - sin^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = 2, b = 1, c = sin^{2} (x)

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

数字相乘:

2 \cdot 1 = 2

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

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

乘开

2 sin (x) (1 - sin^{2} (x)) : 2 sin (x) - 2 sin^{3} (x)

2 sin (x) (1 - sin^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = 2 sin (x), b = 1, c = sin^{2} (x)

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

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

化简

2 \cdot 1 \cdot sin (x) - 2 sin^{2} (x) sin (x) : 2 sin (x) - 2 sin^{3} (x)

2 \cdot 1 \cdot sin (x) - 2 sin^{2} (x) sin (x)

2 \cdot 1 \cdot sin (x) = 2 sin (x)

2 \cdot 1 \cdot sin (x)

数字相乘:

2 \cdot 1 = 2

= 2 sin (x)

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

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

使用指数法则:

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

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

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

数字相加:

2 + 1 = 3

= 2 sin^{3} (x)

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

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

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

化简

- 1 - sin (x) + 2 - 2 sin^{2} (x) + 2 sin (x) - 2 sin^{3} (x) : sin (x) - 2 sin^{3} (x) - 2 sin^{2} (x) + 1

- 1 - sin (x) + 2 - 2 sin^{2} (x) + 2 sin (x) - 2 sin^{3} (x)

对同类项分组

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

同类项相加:

- sin (x) + 2 sin (x) = sin (x)

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

数字相加/相减:

- 1 + 2 = 1

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

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

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

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

用替代法求解

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

令

: sin (x) = u

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

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

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

改写成标准形式

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

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

因式分解

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

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

因式分解出通项

- 1

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

分解

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

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

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

从

- u - 1

分解出因式

- 1

:

- (u + 1)

- u - 1

因式分解出通项

- 1

= - (u + 1)

从

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

分解出因式

2 u^{2}

:

2 u^{2} (u + 1)

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

使用指数法则:

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

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

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

因式分解出通项

2 u^{2}

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

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

因式分解出通项

u + 1

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

分解

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

2 u^{2} - 1

将

2 u^{2} - 1

改写为

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

2 u^{2} - 1

使用根式运算法则:

a = (a)^{2}

2 = (2)^{2}

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

将

1

改写为

1^{2}

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

使用指数法则:

a^{m} b^{m} = (ab)^{m}

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

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

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

使用平方差公式:

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

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

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

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

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

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

使用零因数法则： If

ab = 0

then

a = 0

or

b = 0

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

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

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

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

化简

\frac{2 u}{2} : u

\frac{2 u}{2}

约分:

2

= u

化简

\frac{- 1}{2} : - \frac{2}{2}

\frac{- 1}{2}

使用分式法则:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{1}{2}

- \frac{1}{2}

有理化:

- \frac{2}{2}

- \frac{1}{2}

乘以共轭根式

\frac{2}{2}

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

1 \cdot 2 = 2

22 = 2

22

使用根式运算法则:

a a = a

22 = 2

= 2

= - \frac{2}{2}

= - \frac{2}{2}

u = - \frac{2}{2}

u = - \frac{2}{2}

u = - \frac{2}{2}

解

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

2 u - 1 = 0

将

1

到右边

2 u - 1 = 0

两边加上

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

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

化简

\frac{2 u}{2} : u

\frac{2 u}{2}

约分:

2

= u

化简

\frac{1}{2} : \frac{2}{2}

\frac{1}{2}

乘以共轭根式

\frac{2}{2}

= \frac{1 \cdot 2}{2 2}

1 \cdot 2 = 2

22 = 2

22

使用根式运算法则:

a a = a

22 = 2

= 2

= \frac{2}{2}

u = \frac{2}{2}

u = \frac{2}{2}

u = \frac{2}{2}

解为

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

u = sin (x)

代回

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

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

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

sin (x) = - 1

sin (x) = - 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}

x = \frac{3 π}{2} + 2 πn

x = \frac{3 π}{2} + 2 πn

sin (x) = - \frac{2}{2} : x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

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

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

x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

sin (x) = \frac{2}{2} : x = \frac{π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn

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

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

x = \frac{π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn

x = \frac{π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn

合并所有解

x = \frac{3 π}{2} + 2 πn, x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn, x = \frac{π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn

作图

流行的例子

tan^{2} (b) = 3

solvefor a,sin(a+b/2)=cos(c/2)

so l v e f or a, sin (a + \frac{b}{2}) = cos (\frac{c}{2})

tan(x)-cot(x)=2cot^2(x)

tan (x) - cot (x) = 2 cot^{2} (x)

cot((-x+3n)/4)=0

cot (\frac{- x + 3 n}{4}) = 0

sin(x)+sin^2(x)+sin^3(x)=1

sin (x) + sin^{2} (x) + sin^{3} (x) = 1