zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的三角函数 >

1+sin(2x)=5(sin(x)+cos(x))

解答

1 + sin (2 x) = 5 (sin (x) + cos (x))

解答

x = \frac{3 π}{4} + πn

+1

度数

x = 13 5^{\circ} + 18 0^{\circ} n

求解步骤

1 + sin (2 x) = 5 (sin (x) + cos (x))

两边减去

5 (sin (x) + cos (x))

1 + sin (2 x) - 5 (sin (x) + cos (x)) = 0

使用三角恒等式改写

1 + sin (2 x) - (cos (x) + sin (x)) \cdot 5

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

1 + sin (2 x)

使用毕达哥拉斯恒等式:

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

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

使用倍角公式:

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

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

使用完全平方公式:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

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

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

= (sin (x) + cos (x))^{2} - 5 (cos (x) + sin (x))

(cos (x) + sin (x))^{2} - (cos (x) + sin (x)) \cdot 5 = 0

分解

(cos (x) + sin (x))^{2} - (cos (x) + sin (x)) \cdot 5 : (cos (x) + sin (x)) (cos (x) + sin (x) - 5)

(cos (x) + sin (x))^{2} - (cos (x) + sin (x)) \cdot 5

因式分解出通项

(cos (x) + sin (x)) : (cos (x) + sin (x)) ((cos (x) + sin (x)) - 5)

(cos (x) + sin (x))^{2} - 5 (cos (x) + sin (x))

使用指数法则:

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

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

= (cos (x) + sin (x)) (cos (x) + sin (x)) - 5 (cos (x) + sin (x))

因式分解出通项

(cos (x) + sin (x))

= (cos (x) + sin (x)) ((cos (x) + sin (x)) - 5)

= (cos (x) + sin (x)) ((cos (x) + sin (x)) - 5)

整理后得

= (cos (x) + sin (x)) (cos (x) + sin (x) - 5)

(cos (x) + sin (x)) (cos (x) + sin (x) - 5) = 0

分别求解每个部分

cos (x) + sin (x) = 0 or cos (x) + sin (x) - 5 = 0

cos (x) + sin (x) = 0 : x = \frac{3 π}{4} + πn

cos (x) + sin (x) = 0

使用三角恒等式改写

cos (x) + sin (x) = 0

在两边除以

cos (x), cos (x) \neq = 0

\frac{cos ( x ) + sin ( x )}{cos ( x )} = \frac{0}{cos ( x )}

化简

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

使用基本三角恒等式:

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

1 + tan (x) = 0

1 + tan (x) = 0

将

1

到右边

1 + tan (x) = 0

两边减去

1

1 + tan (x) - 1 = 0 - 1

化简

tan (x) = - 1

tan (x) = - 1

tan (x) = - 1

的通解

tan (x)

周期表（周期为

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

cos (x) + sin (x) - 5 = 0 :

无解

cos (x) + sin (x) - 5 = 0

使用三角恒等式改写

cos (x) + sin (x) - 5

sin (x) + cos (x) = 2 sin (x + \frac{π}{4})

sin (x) + cos (x)

改写为

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

使用以下普通恒等式:

cos (\frac{π}{4}) = \frac{1}{2}

使用以下普通恒等式:

sin (\frac{π}{4}) = \frac{1}{2}

= 2 (cos (\frac{π}{4}) sin (x) + sin (\frac{π}{4}) cos (x))

使用角和恒等式:

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

= 2 sin (x + \frac{π}{4})

= - 5 + 2 sin (x + \frac{π}{4})

- 5 + 2 sin (x + \frac{π}{4}) = 0

将

5

到右边

- 5 + 2 sin (x + \frac{π}{4}) = 0

两边加上

5

- 5 + 2 sin (x + \frac{π}{4}) + 5 = 0 + 5

化简

2 sin (x + \frac{π}{4}) = 5

2 sin (x + \frac{π}{4}) = 5

两边除以

2

2 sin (x + \frac{π}{4}) = 5

两边除以

2

\frac{2 sin ( x + \frac{π}{4} )}{2} = \frac{5}{2}

化简

\frac{2 sin ( x + \frac{π}{4} )}{2} = \frac{5}{2}

化简

\frac{2 sin ( x + \frac{π}{4} )}{2} : sin (x + \frac{π}{4})

\frac{2 sin ( x + \frac{π}{4} )}{2}

约分:

2

= sin (x + \frac{π}{4})

化简

\frac{5}{2} : \frac{5 2}{2}

\frac{5}{2}

乘以共轭根式

\frac{2}{2}

= \frac{5 2}{2 2}

22 = 2

22

使用根式运算法则:

a a = a

22 = 2

= 2

= \frac{5 2}{2}

sin (x + \frac{π}{4}) = \frac{5 2}{2}

sin (x + \frac{π}{4}) = \frac{5 2}{2}

sin (x + \frac{π}{4}) = \frac{5 2}{2}

- 1 \leq sin (x) \leq 1

无解

合并所有解

x = \frac{3 π}{4} + πn

作图

流行的例子

cos(2x)= 3/4-5cos^2(x)

cos (2 x) = \frac{3}{4} - 5 cos^{2} (x)

sin(x+pi/9)=(sqrt(3))/2

sin (x + \frac{π}{9}) = \frac{3}{2}

3cos(2θ)=3sin(θ)

3 cos (2 θ) = 3 sin (θ)

-sec(θ)cot(θ)-sec(θ)=-2sec(θ)

- sec (θ) cot (θ) - sec (θ) = - 2 sec (θ)

sin(5x-27)=cos(6x+13.6)

sin (5 x - 27) = cos (6 x + 13.6)