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受欢迎的三角函数 >

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

解答

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

解答

x = \frac{π}{4} + πn

+1

度数

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

求解步骤

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

两边减去

2

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

使用三角恒等式改写

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

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

化简

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

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

用替代法求解

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

令

: sin (2 x) = u

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

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

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

去除平方根

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

两边减去

u

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

化简

2 - u^{2} - 2 = - u

两边加上

2

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

化简

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

两边进行平方:

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

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

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

展开

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

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

使用根式运算法则:

a = a^{\frac{1}{2}}

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

使用指数法则:

(a^{b})^{c} = a^{b c}

= (2 - u^{2})^{\frac{1}{2} \cdot 2}

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

\frac{1}{2} \cdot 2

分式相乘:

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

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

约分:

2

= 1

= 2 - u^{2}

展开

(- u + 2)^{2} : u^{2} - 4 u + 4

(- u + 2)^{2}

使用完全平方公式:

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

a = - u, b = 2

= (- u)^{2} + 2 (- u) \cdot 2 + 2^{2}

化简

(- u)^{2} + 2 (- u) \cdot 2 + 2^{2} : u^{2} - 4 u + 4

(- u)^{2} + 2 (- u) \cdot 2 + 2^{2}

去除括号:

(- a) = - a

= (- u)^{2} - 2 u \cdot 2 + 2^{2}

使用指数法则:

(- a)^{n} = a^{n},

若

n

是偶数

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

= u^{2} - 2 \cdot 2 u + 2^{2}

整理后得

= u^{2} - 4 u + 4

= u^{2} - 4 u + 4

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

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

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

解

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

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

交换两边

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

将

u^{2}

para o lado esquerdo

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

两边加上

u^{2}

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

化简

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

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

将

2

para o lado esquerdo

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

两边减去

2

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

化简

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

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

使用求根公式求解

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

二次方程求根公式:

若

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

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

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

(- 4)^{2} - 4 \cdot 2 \cdot 2 = 0

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

使用指数法则:

(- a)^{n} = a^{n},

若

n

是偶数

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

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

数字相乘:

4 \cdot 2 \cdot 2 = 16

= 4^{2} - 16

4^{2} = 16

= 16 - 16

数字相减:

16 - 16 = 0

= 0

u_{1, 2} = \frac{- ( - 4 ) \pm 0}{2 \cdot 2}

u = \frac{- ( - 4 )}{2 \cdot 2}

\frac{- ( - 4 )}{2 \cdot 2} = 1

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

使用法则

- (- a) = a

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

数字相乘:

2 \cdot 2 = 4

= \frac{4}{4}

使用法则

\frac{a}{a} = 1

= 1

u = 1

二次方程组的解是:

u = 1

u = 1

验证解:

u = 1

真

将它们代入

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

检验解是否符合

去除与方程不符的解。

代入

u = 1 :

真

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

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

- 2 + 1 + 2 - 1^{2}

使用法则

1^{a} = 1

1^{2} = 1

= - 2 + 1 + 2 - 1

2 - 1 = 1

2 - 1

数字相减:

2 - 1 = 1

= 1

使用法则

1 = 1

= 1

= - 2 + 1 + 1

数字相加/相减:

- 2 + 1 + 1 = 0

= 0

0 = 0

真

解是

u = 1

u = sin (2 x)

代回

sin (2 x) = 1

sin (2 x) = 1

sin (2 x) = 1 : x = \frac{π}{4} + πn

sin (2 x) = 1

sin (2 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}

2 x = \frac{π}{2} + 2 πn

2 x = \frac{π}{2} + 2 πn

解

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

2 x = \frac{π}{2} + 2 πn

两边除以

2

2 x = \frac{π}{2} + 2 πn

两边除以

2

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

化简

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

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

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

\frac{\frac{π}{2}}{2} + \frac{2 πn}{2}

\frac{\frac{π}{2}}{2} = \frac{π}{4}

\frac{\frac{π}{2}}{2}

使用分式法则:

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

= \frac{π}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{π}{4}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

= \frac{π}{4} + πn

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

合并所有解

x = \frac{π}{4} + πn

作图

流行的例子

8sin^2(x)2cos(x)=7,0<= x<= 2pi

8 sin^{2} (x) 2 cos (x) = 7, 0 \leq x \leq 2 π

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

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

9cos(2x)=9cos^2(x)-4

9 cos (2 x) = 9 cos^{2} (x) - 4

2 - cos (x) = 0

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