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受欢迎的三角函数 >

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

解答

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

解答

x \in R 无解

求解步骤

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

两边减去

5

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

使用三角恒等式改写

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

= - 5 + sin (x) + 2 (1 - sin^{2} (x))

化简

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

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

乘开

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

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

使用分配律:

a (b - c) = ab - a c

a = 2, b = 1, c = sin^{2} (x)

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

数字相乘:

2 \cdot 1 = 2

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

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

化简

- 5 + sin (x) + 2 - 2 sin^{2} (x) : sin (x) - 2 sin^{2} (x) - 3

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

对同类项分组

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

数字相加/相减:

- 5 + 2 = - 3

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

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

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

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

用替代法求解

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

令

: sin (x) = u

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

- 3 + u - 2 u^{2} = 0 : u = \frac{1}{4} - i \frac{23}{4}, u = \frac{1}{4} + i \frac{23}{4}

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

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

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

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

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

化简

1^{2} - 4 (- 2) (- 3) : 23 i

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

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 2) (- 3)

使用法则

- (- a) = a

= 1 - 4 \cdot 2 \cdot 3

数字相乘:

4 \cdot 2 \cdot 3 = 24

= 1 - 24

数字相减:

1 - 24 = - 23

= - 23

使用根式运算法则:

- a = - 1 a

- 23 = - 1 23

= - 1 23

使用虚数运算法则:

- 1 = i

= 23 i

u_{1, 2} = \frac{- 1 \pm 23 i}{2 ( - 2 )}

将解分隔开

u_{1} = \frac{- 1 + 23 i}{2 ( - 2 )}, u_{2} = \frac{- 1 - 23 i}{2 ( - 2 )}

u = \frac{- 1 + 23 i}{2 ( - 2 )} : \frac{1}{4} - i \frac{23}{4}

\frac{- 1 + 23 i}{2 ( - 2 )}

去除括号:

(- a) = - a

= \frac{- 1 + 23 i}{- 2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{- 1 + 23 i}{- 4}

使用分式法则:

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

= - \frac{- 1 + 23 i}{4}

将

- \frac{- 1 + 23 i}{4}

改写成标准复数形式:

\frac{1}{4} - \frac{23}{4} i

- \frac{- 1 + 23 i}{4}

使用分式法则:

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

\frac{- 1 + 23 i}{4} = - (- \frac{1}{4}) - (\frac{23 i}{4})

= - (- \frac{1}{4}) - (\frac{23 i}{4})

去除括号:

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

= \frac{1}{4} - \frac{23 i}{4}

= \frac{1}{4} - \frac{23}{4} i

u = \frac{- 1 - 23 i}{2 ( - 2 )} : \frac{1}{4} + i \frac{23}{4}

\frac{- 1 - 23 i}{2 ( - 2 )}

去除括号:

(- a) = - a

= \frac{- 1 - 23 i}{- 2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{- 1 - 23 i}{- 4}

使用分式法则:

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

= - \frac{- 1 - 23 i}{4}

将

- \frac{- 1 - 23 i}{4}

改写成标准复数形式:

\frac{1}{4} + \frac{23}{4} i

- \frac{- 1 - 23 i}{4}

使用分式法则:

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

\frac{- 1 - 23 i}{4} = - (- \frac{1}{4}) - (- \frac{23 i}{4})

= - (- \frac{1}{4}) - (- \frac{23 i}{4})

使用法则

- (- a) = a

= \frac{1}{4} + \frac{23 i}{4}

= \frac{1}{4} + \frac{23}{4} i

二次方程组的解是:

u = \frac{1}{4} - i \frac{23}{4}, u = \frac{1}{4} + i \frac{23}{4}

u = sin (x)

代回

sin (x) = \frac{1}{4} - i \frac{23}{4}, sin (x) = \frac{1}{4} + i \frac{23}{4}

sin (x) = \frac{1}{4} - i \frac{23}{4}, sin (x) = \frac{1}{4} + i \frac{23}{4}

sin (x) = \frac{1}{4} - i \frac{23}{4} :

无解

sin (x) = \frac{1}{4} - i \frac{23}{4}

无解

sin (x) = \frac{1}{4} + i \frac{23}{4} :

无解

sin (x) = \frac{1}{4} + i \frac{23}{4}

无解

合并所有解

x \in R 无解

作图

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