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受欢迎的三角函数 >

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

解答

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

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

使用三角恒等式改写

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

化简

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

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

用替代法求解

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

令

: sin (2 x) = u

- 2 - u^{2} + 3 u = 0

- 2 - u^{2} + 3 u = 0 : u = 1, u = 2

- 2 - u^{2} + 3 u = 0

改写成标准形式

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

- u^{2} + 3 u - 2 = 0

使用求根公式求解

- u^{2} + 3 u - 2 = 0

二次方程求根公式:

若

a = - 1, b = 3, c = - 2

u_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 ( - 1 ) ( - 2 )}{2 ( - 1 )}

u_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 ( - 1 ) ( - 2 )}{2 ( - 1 )}

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

3^{2} - 4 (- 1) (- 2)

使用法则

- (- a) = a

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

数字相乘:

4 \cdot 1 \cdot 2 = 8

= 3^{2} - 8

3^{2} = 9

= 9 - 8

数字相减:

9 - 8 = 1

= 1

使用法则

1 = 1

= 1

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

将解分隔开

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

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

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

去除括号:

(- a) = - a

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

数字相加/相减:

- 3 + 1 = - 2

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

数字相乘:

2 \cdot 1 = 2

= \frac{- 2}{- 2}

使用分式法则:

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

= \frac{2}{2}

使用法则

\frac{a}{a} = 1

= 1

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

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

去除括号:

(- a) = - a

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

数字相减:

- 3 - 1 = - 4

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

数字相乘:

2 \cdot 1 = 2

= \frac{- 4}{- 2}

使用分式法则:

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

= \frac{4}{2}

数字相除:

\frac{4}{2} = 2

= 2

二次方程组的解是:

u = 1, u = 2

u = sin (2 x)

代回

sin (2 x) = 1, sin (2 x) = 2

sin (2 x) = 1, sin (2 x) = 2

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

sin (2 x) = 2 :

无解

sin (2 x) = 2

- 1 \leq sin (x) \leq 1

无解

合并所有解

x = \frac{π}{4} + πn

作图

流行的例子

sqrt(1-cos^2(x))=0.75sqrt(1-cos^2(y))

1 - cos^{2} (x) = 0.75 1 - cos^{2} (y)

sec^2(x)-6tan(x)+4=0

sec^{2} (x) - 6 tan (x) + 4 = 0

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

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

-5tan(x)-12=tan(x)-8

- 5 tan (x) - 12 = tan (x) - 8

5 = sec^{2} (x) + 3