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

解答

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

解答

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn, x = \frac{π}{2} + 2 πn, x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = \frac{3 π}{2} + 2 πn

+1

度数

x = 21 0^{\circ} + 36 0^{\circ} n, x = 33 0^{\circ} + 36 0^{\circ} n, x = 9 0^{\circ} + 36 0^{\circ} n, x = 3 0^{\circ} + 36 0^{\circ} n, x = 15 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n

求解步骤

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

两边减去

1

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

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

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

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

分解

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

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

使用平方差公式:

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

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

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

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

分别求解每个部分

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

sin (x) + cos (2 x) = 0 : x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn, x = \frac{π}{2} + 2 πn

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

使用三角恒等式改写

cos (2 x) + sin (x)

使用倍角公式:

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

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

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

用替代法求解

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

令

: sin (x) = u

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

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

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

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

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

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

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

1^{2} - 4 (- 2) \cdot 1 = 3

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

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 2) \cdot 1

使用法则

- (- a) = a

= 1 + 4 \cdot 2 \cdot 1

数字相乘:

4 \cdot 2 \cdot 1 = 8

= 1 + 8

数字相加:

1 + 8 = 9

= 9

因式分解数字:

9 = 3^{2}

= 3^{2}

使用根式运算法则:

3^{2} = 3

= 3

u_{1, 2} = \frac{- 1 \pm 3}{2 ( - 2 )}

将解分隔开

u_{1} = \frac{- 1 + 3}{2 ( - 2 )}, u_{2} = \frac{- 1 - 3}{2 ( - 2 )}

u = \frac{- 1 + 3}{2 ( - 2 )} : - \frac{1}{2}

\frac{- 1 + 3}{2 ( - 2 )}

去除括号:

(- a) = - a

= \frac{- 1 + 3}{- 2 \cdot 2}

数字相加/相减:

- 1 + 3 = 2

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

数字相乘:

2 \cdot 2 = 4

= \frac{2}{- 4}

使用分式法则:

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

= - \frac{2}{4}

约分:

2

= - \frac{1}{2}

u = \frac{- 1 - 3}{2 ( - 2 )} : 1

\frac{- 1 - 3}{2 ( - 2 )}

去除括号:

(- a) = - a

= \frac{- 1 - 3}{- 2 \cdot 2}

数字相减:

- 1 - 3 = - 4

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

数字相乘:

2 \cdot 2 = 4

= \frac{- 4}{- 4}

使用分式法则:

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

= \frac{4}{4}

使用法则

\frac{a}{a} = 1

= 1

二次方程组的解是:

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

u = sin (x)

代回

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

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

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

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

sin (x) = - \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}

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

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

sin (x) = 1 : x = \frac{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{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

合并所有解

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

sin (x) - cos (2 x) = 0 : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = \frac{3 π}{2} + 2 πn

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

使用三角恒等式改写

- cos (2 x) + sin (x)

使用倍角公式:

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

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

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

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

打开括号

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

使用加减运算法则

- (- a) = a, - (a) = - a

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

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

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

用替代法求解

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

令

: sin (x) = u

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

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

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

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

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

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

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

1^{2} - 4 \cdot 2 (- 1) = 3

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

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 2 (- 1)

使用法则

- (- a) = a

= 1 + 4 \cdot 2 \cdot 1

数字相乘:

4 \cdot 2 \cdot 1 = 8

= 1 + 8

数字相加:

1 + 8 = 9

= 9

因式分解数字:

9 = 3^{2}

= 3^{2}

使用根式运算法则:

3^{2} = 3

= 3

u_{1, 2} = \frac{- 1 \pm 3}{2 \cdot 2}

将解分隔开

u_{1} = \frac{- 1 + 3}{2 \cdot 2}, u_{2} = \frac{- 1 - 3}{2 \cdot 2}

u = \frac{- 1 + 3}{2 \cdot 2} : \frac{1}{2}

\frac{- 1 + 3}{2 \cdot 2}

数字相加/相减:

- 1 + 3 = 2

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

数字相乘:

2 \cdot 2 = 4

= \frac{2}{4}

约分:

2

= \frac{1}{2}

u = \frac{- 1 - 3}{2 \cdot 2} : - 1

\frac{- 1 - 3}{2 \cdot 2}

数字相减:

- 1 - 3 = - 4

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

数字相乘:

2 \cdot 2 = 4

= \frac{- 4}{4}

使用分式法则:

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

= - \frac{4}{4}

使用法则

\frac{a}{a} = 1

= - 1

二次方程组的解是:

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

u = sin (x)

代回

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

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

sin (x) = \frac{1}{2} : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

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

sin (x) = \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}

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

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

合并所有解

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = \frac{3 π}{2} + 2 πn

合并所有解

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn, x = \frac{π}{2} + 2 πn, x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = \frac{3 π}{2} + 2 πn

流行的例子

tan (2 x - 5) = 1

5/(sqrt(21)*\sqrt{30)}=cos(x)

\frac{5}{21 \cdot 30} = cos (x)

cos^2(x)=((7k-7))/3

cos^{2} (x) = \frac{( 7 k - 7 )}{3}

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

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

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

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