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受欢迎的三角函数 >

sin(2x)-sqrt(2)cos(x)=sqrt(2)sin(x)-1

解答

sin (2 x) - 2 cos (x) = 2 sin (x) - 1

解答

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

+1

度数

x = 13 5^{\circ} + 18 0^{\circ} n, x = 4 5^{\circ} + 36 0^{\circ} n

求解步骤

sin (2 x) - 2 cos (x) = 2 sin (x) - 1

两边减去

2 sin (x) - 1

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

使用三角恒等式改写

1 + sin (2 x) - cos (x) 2 - sin (x) 2

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

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

分解

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

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

因式分解出通项

2

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

因式分解出通项

(cos (x) + sin (x))

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

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

分别求解每个部分

cos (x) + sin (x) = 0 or cos (x) + sin (x) - 2 = 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) - 2 = 0 : x = 2 πn + \frac{π}{4}

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

使用三角恒等式改写

cos (x) + sin (x) - 2

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})

= - 2 + 2 sin (x + \frac{π}{4})

- 2 + 2 sin (x + \frac{π}{4}) = 0

将

2

到右边

- 2 + 2 sin (x + \frac{π}{4}) = 0

两边加上

2

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

化简

2 sin (x + \frac{π}{4}) = 2

2 sin (x + \frac{π}{4}) = 2

两边除以

2

2 sin (x + \frac{π}{4}) = 2

两边除以

2

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

化简

sin (x + \frac{π}{4}) = 1

sin (x + \frac{π}{4}) = 1

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

x + \frac{π}{4} = \frac{π}{2} + 2 πn

解

x + \frac{π}{4} = \frac{π}{2} + 2 πn : x = 2 πn + \frac{π}{4}

x + \frac{π}{4} = \frac{π}{2} + 2 πn

将

\frac{π}{4}

到右边

x + \frac{π}{4} = \frac{π}{2} + 2 πn

两边减去

\frac{π}{4}

x + \frac{π}{4} - \frac{π}{4} = \frac{π}{2} + 2 πn - \frac{π}{4}

化简

x + \frac{π}{4} - \frac{π}{4} = \frac{π}{2} + 2 πn - \frac{π}{4}

化简

x + \frac{π}{4} - \frac{π}{4} : x

x + \frac{π}{4} - \frac{π}{4}

同类项相加:

\frac{π}{4} - \frac{π}{4} = 0

= x

化简

\frac{π}{2} + 2 πn - \frac{π}{4} : 2 πn + \frac{π}{4}

\frac{π}{2} + 2 πn - \frac{π}{4}

对同类项分组

= 2 πn + \frac{π}{2} - \frac{π}{4}

2, 4

的最小公倍数:

4

2, 4

最小公倍数 (LCM)

2

质因数分解:

2

2

2

是质数，因此无法因数分解

= 2

4

质因数分解:

2 \cdot 2

4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2

将每个因子乘以它在

2

或

4

中出现的最多次数

= 2 \cdot 2

数字相乘:

2 \cdot 2 = 4

= 4

根据最小公倍数调整分式

将每个分子乘以其分母转变为最小公倍数所要乘以的同一数值

4

对于

\frac{π}{2} :

将分母和分子乘以

2

\frac{π}{2} = \frac{π 2}{2 \cdot 2} = \frac{π 2}{4}

= \frac{π 2}{4} - \frac{π}{4}

因为分母相等，所以合并分式:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{π 2 - π}{4}

同类项相加:

2 π - π = π

= 2 πn + \frac{π}{4}

x = 2 πn + \frac{π}{4}

x = 2 πn + \frac{π}{4}

x = 2 πn + \frac{π}{4}

x = 2 πn + \frac{π}{4}

合并所有解

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

作图

流行的例子

cos^2(θ)-3/2 cos(2θ)=0

cos^{2} (θ) - \frac{3}{2} cos (2 θ) = 0

sinh(x)= 3/4 ,cosh(x)

sinh (x) = \frac{3}{4}, cosh (x)

tan(θ)=1.4739716

tan (θ) = 1.4739716

cos(2x)=cos(1/2)

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

cos (x) = \frac{3}{10}