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受欢迎的三角函数 >

sin(2x)+cos(3x)=0

解答

sin (2 x) + cos (3 x) = 0

解答

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = - 0.31415 \dots + 2 πn, x = π + 0.31415 \dots + 2 πn, x = 0.94247 \dots + 2 πn, x = π - 0.94247 \dots + 2 πn

+1

度数

x = 9 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n, x = - 1 8^{\circ} + 36 0^{\circ} n, x = 19 8^{\circ} + 36 0^{\circ} n, x = 5 4^{\circ} + 36 0^{\circ} n, x = 12 6^{\circ} + 36 0^{\circ} n

求解步骤

sin (2 x) + cos (3 x) = 0

使用三角恒等式改写

cos (3 x) + sin (2 x)

使用倍角公式:

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

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

cos (3 x) = 4 cos^{3} (x) - 3 cos (x)

cos (3 x)

使用三角恒等式改写

cos (3 x)

改写为

= cos (2 x + x)

使用角和恒等式:

cos (s + t) = cos (s) cos (t) - sin (s) sin (t)

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

使用倍角公式:

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

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

化简

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

cos (2 x) cos (x) - 2 sin (x) cos (x) sin (x)

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

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

使用指数法则:

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

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

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

数字相加:

1 + 1 = 2

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

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

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

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

使用倍角公式:

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

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

使用毕达哥拉斯恒等式:

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

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

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

乘开

(2 cos^{2} (x) - 1) cos (x) - 2 (1 - cos^{2} (x)) cos (x) : 4 cos^{3} (x) - 3 cos (x)

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

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

乘开

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

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

使用分配律:

a (b - c) = ab - a c

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

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

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

化简

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

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

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

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

使用指数法则:

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

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

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

数字相加:

2 + 1 = 3

= 2 cos^{3} (x)

1 \cdot cos (x) = cos (x)

1 cos (x)

乘以:

1 \cdot cos (x) = cos (x)

= cos (x)

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

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

= 2 cos^{3} (x) - cos (x) - 2 (1 - cos^{2} (x)) cos (x)

乘开

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

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

使用分配律:

a (b - c) = ab - a c

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

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

使用加减运算法则

- (- a) = a

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

化简

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

- 2 \cdot 1 cos (x) + 2 cos^{2} (x) cos (x)

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

2 \cdot 1 cos (x)

数字相乘:

2 \cdot 1 = 2

= 2 cos (x)

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

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

使用指数法则:

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

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

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

数字相加:

2 + 1 = 3

= 2 cos^{3} (x)

= - 2 cos (x) + 2 cos^{3} (x)

= - 2 cos (x) + 2 cos^{3} (x)

= 2 cos^{3} (x) - cos (x) - 2 cos (x) + 2 cos^{3} (x)

化简

2 cos^{3} (x) - cos (x) - 2 cos (x) + 2 cos^{3} (x) : 4 cos^{3} (x) - 3 cos (x)

2 cos^{3} (x) - cos (x) - 2 cos (x) + 2 cos^{3} (x)

对同类项分组

= 2 cos^{3} (x) + 2 cos^{3} (x) - cos (x) - 2 cos (x)

同类项相加:

2 cos^{3} (x) + 2 cos^{3} (x) = 4 cos^{3} (x)

= 4 cos^{3} (x) - cos (x) - 2 cos (x)

同类项相加:

- cos (x) - 2 cos (x) = - 3 cos (x)

= 4 cos^{3} (x) - 3 cos (x)

= 4 cos^{3} (x) - 3 cos (x)

= 4 cos^{3} (x) - 3 cos (x)

= 4 cos^{3} (x) - 3 cos (x) + 2 cos (x) sin (x)

- 3 cos (x) + 4 cos^{3} (x) + 2 cos (x) sin (x) = 0

分解

- 3 cos (x) + 4 cos^{3} (x) + 2 cos (x) sin (x) : cos (x) (- 3 + 4 cos^{2} (x) + 2 sin (x))

- 3 cos (x) + 4 cos^{3} (x) + 2 cos (x) sin (x)

使用指数法则:

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

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

= - 3 cos (x) + 4 cos (x) cos^{2} (x) + 2 sin (x) cos (x)

因式分解出通项

cos (x)

= cos (x) (- 3 + 4 cos^{2} (x) + 2 sin (x))

cos (x) (- 3 + 4 cos^{2} (x) + 2 sin (x)) = 0

分别求解每个部分

cos (x) = 0 or - 3 + 4 cos^{2} (x) + 2 sin (x) = 0

cos (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos (x) = 0

cos (x) = 0

的通解

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

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

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

- 3 + 4 cos^{2} (x) + 2 sin (x) = 0 : x = arcsin (- \frac{- 1 + 5}{4}) + 2 πn, x = π + arcsin (\frac{- 1 + 5}{4}) + 2 πn, x = arcsin (\frac{1 + 5}{4}) + 2 πn, x = π - arcsin (\frac{1 + 5}{4}) + 2 πn

- 3 + 4 cos^{2} (x) + 2 sin (x) = 0

使用三角恒等式改写

- 3 + 2 sin (x) + 4 cos^{2} (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)

= - 3 + 2 sin (x) + 4 (1 - sin^{2} (x))

化简

- 3 + 2 sin (x) + 4 (1 - sin^{2} (x)) : 2 sin (x) - 4 sin^{2} (x) + 1

- 3 + 2 sin (x) + 4 (1 - sin^{2} (x))

乘开

4 (1 - sin^{2} (x)) : 4 - 4 sin^{2} (x)

4 (1 - sin^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = 4, b = 1, c = sin^{2} (x)

= 4 \cdot 1 - 4 sin^{2} (x)

数字相乘:

4 \cdot 1 = 4

= 4 - 4 sin^{2} (x)

= - 3 + 2 sin (x) + 4 - 4 sin^{2} (x)

化简

- 3 + 2 sin (x) + 4 - 4 sin^{2} (x) : 2 sin (x) - 4 sin^{2} (x) + 1

- 3 + 2 sin (x) + 4 - 4 sin^{2} (x)

对同类项分组

= 2 sin (x) - 4 sin^{2} (x) - 3 + 4

数字相加/相减:

- 3 + 4 = 1

= 2 sin (x) - 4 sin^{2} (x) + 1

= 2 sin (x) - 4 sin^{2} (x) + 1

= 2 sin (x) - 4 sin^{2} (x) + 1

1 + 2 sin (x) - 4 sin^{2} (x) = 0

用替代法求解

1 + 2 sin (x) - 4 sin^{2} (x) = 0

令

: sin (x) = u

1 + 2 u - 4 u^{2} = 0

1 + 2 u - 4 u^{2} = 0 : u = - \frac{- 1 + 5}{4}, u = \frac{1 + 5}{4}

1 + 2 u - 4 u^{2} = 0

改写成标准形式

a x^{2} + b x + c = 0

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

使用求根公式求解

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

二次方程求根公式:

若

a = - 4, b = 2, c = 1

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 ( - 4 ) \cdot 1}{2 ( - 4 )}

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 ( - 4 ) \cdot 1}{2 ( - 4 )}

2^{2} - 4 (- 4) \cdot 1 = 25

2^{2} - 4 (- 4) \cdot 1

使用法则

- (- a) = a

= 2^{2} + 4 \cdot 4 \cdot 1

数字相乘:

4 \cdot 4 \cdot 1 = 16

= 2^{2} + 16

2^{2} = 4

= 4 + 16

数字相加:

4 + 16 = 20

= 20

20

质因数分解:

2^{2} \cdot 5

20

20

除以

220 = 10 \cdot 2

= 2 \cdot 10

10

除以

210 = 5 \cdot 2

= 2 \cdot 2 \cdot 5

2, 5

都是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 5

= 2^{2} \cdot 5

= 2^{2} \cdot 5

使用根式运算法则:

n ab = n a n b

= 5 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 25

u_{1, 2} = \frac{- 2 \pm 2 5}{2 ( - 4 )}

将解分隔开

u_{1} = \frac{- 2 + 2 5}{2 ( - 4 )}, u_{2} = \frac{- 2 - 2 5}{2 ( - 4 )}

u = \frac{- 2 + 2 5}{2 ( - 4 )} : - \frac{- 1 + 5}{4}

\frac{- 2 + 2 5}{2 ( - 4 )}

去除括号:

(- a) = - a

= \frac{- 2 + 2 5}{- 2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{- 2 + 2 5}{- 8}

使用分式法则:

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

= - \frac{- 2 + 2 5}{8}

消掉

\frac{- 2 + 2 5}{8} : \frac{5 - 1}{4}

\frac{- 2 + 2 5}{8}

分解

- 2 + 25 : 2 (- 1 + 5)

- 2 + 25

改写为

= - 2 \cdot 1 + 25

因式分解出通项

2

= 2 (- 1 + 5)

= \frac{2 ( - 1 + 5 )}{8}

约分:

2

= \frac{- 1 + 5}{4}

= - \frac{5 - 1}{4}

= - \frac{- 1 + 5}{4}

u = \frac{- 2 - 2 5}{2 ( - 4 )} : \frac{1 + 5}{4}

\frac{- 2 - 2 5}{2 ( - 4 )}

去除括号:

(- a) = - a

= \frac{- 2 - 2 5}{- 2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{- 2 - 2 5}{- 8}

使用分式法则:

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

- 2 - 25 = - (2 + 25)

= \frac{2 + 2 5}{8}

分解

2 + 25 : 2 (1 + 5)

2 + 25

改写为

= 2 \cdot 1 + 25

因式分解出通项

2

= 2 (1 + 5)

= \frac{2 ( 1 + 5 )}{8}

约分:

2

= \frac{1 + 5}{4}

二次方程组的解是:

u = - \frac{- 1 + 5}{4}, u = \frac{1 + 5}{4}

u = sin (x)

代回

sin (x) = - \frac{- 1 + 5}{4}, sin (x) = \frac{1 + 5}{4}

sin (x) = - \frac{- 1 + 5}{4}, sin (x) = \frac{1 + 5}{4}

sin (x) = - \frac{- 1 + 5}{4} : x = arcsin (- \frac{- 1 + 5}{4}) + 2 πn, x = π + arcsin (\frac{- 1 + 5}{4}) + 2 πn

sin (x) = - \frac{- 1 + 5}{4}

使用反三角函数性质

sin (x) = - \frac{- 1 + 5}{4}

sin (x) = - \frac{- 1 + 5}{4}

的通解

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

x = arcsin (- \frac{- 1 + 5}{4}) + 2 πn, x = π + arcsin (\frac{- 1 + 5}{4}) + 2 πn

x = arcsin (- \frac{- 1 + 5}{4}) + 2 πn, x = π + arcsin (\frac{- 1 + 5}{4}) + 2 πn

sin (x) = \frac{1 + 5}{4} : x = arcsin (\frac{1 + 5}{4}) + 2 πn, x = π - arcsin (\frac{1 + 5}{4}) + 2 πn

sin (x) = \frac{1 + 5}{4}

使用反三角函数性质

sin (x) = \frac{1 + 5}{4}

sin (x) = \frac{1 + 5}{4}

的通解

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

x = arcsin (\frac{1 + 5}{4}) + 2 πn, x = π - arcsin (\frac{1 + 5}{4}) + 2 πn

x = arcsin (\frac{1 + 5}{4}) + 2 πn, x = π - arcsin (\frac{1 + 5}{4}) + 2 πn

合并所有解

x = arcsin (- \frac{- 1 + 5}{4}) + 2 πn, x = π + arcsin (\frac{- 1 + 5}{4}) + 2 πn, x = arcsin (\frac{1 + 5}{4}) + 2 πn, x = π - arcsin (\frac{1 + 5}{4}) + 2 πn

合并所有解

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = arcsin (- \frac{- 1 + 5}{4}) + 2 πn, x = π + arcsin (\frac{- 1 + 5}{4}) + 2 πn, x = arcsin (\frac{1 + 5}{4}) + 2 πn, x = π - arcsin (\frac{1 + 5}{4}) + 2 πn

以小数形式表示解

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = - 0.31415 \dots + 2 πn, x = π + 0.31415 \dots + 2 πn, x = 0.94247 \dots + 2 πn, x = π - 0.94247 \dots + 2 πn

作图

流行的例子

2cos^2(x)+2cos(x)-1=0,0<= x<= 2pi

2 cos^{2} (x) + 2 cos (x) - 1 = 0, 0 \leq x \leq 2 π

sec (x) = \frac{7}{5}

cos(5θ)=sin(3θ)

cos (5 θ) = sin (3 θ)

(2sin(2x)+2cos(2x))^2=4

(2 sin (2 x) + 2 cos (2 x))^{2} = 4

sin (x) = \frac{7}{3}