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受欢迎的三角函数 >

4sin^2(x)=2cos(x)+1

解答

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

解答

x = 0.86138 \dots + 2 πn, x = 2 π - 0.86138 \dots + 2 πn

+1

度数

x = 49.35368 \dots^{\circ} + 36 0^{\circ} n, x = 310.64631 \dots^{\circ} + 36 0^{\circ} n

求解步骤

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

两边减去

2 cos (x) + 1

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

使用三角恒等式改写

- 1 - 2 cos (x) + 4 sin^{2} (x)

使用毕达哥拉斯恒等式:

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

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

= - 1 - 2 cos (x) + 4 (1 - cos^{2} (x))

化简

- 1 - 2 cos (x) + 4 (1 - cos^{2} (x)) : - 4 cos^{2} (x) - 2 cos (x) + 3

- 1 - 2 cos (x) + 4 (1 - cos^{2} (x))

乘开

4 (1 - cos^{2} (x)) : 4 - 4 cos^{2} (x)

4 (1 - cos^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = 4, b = 1, c = cos^{2} (x)

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

数字相乘:

4 \cdot 1 = 4

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

= - 1 - 2 cos (x) + 4 - 4 cos^{2} (x)

化简

- 1 - 2 cos (x) + 4 - 4 cos^{2} (x) : - 4 cos^{2} (x) - 2 cos (x) + 3

- 1 - 2 cos (x) + 4 - 4 cos^{2} (x)

对同类项分组

= - 2 cos (x) - 4 cos^{2} (x) - 1 + 4

数字相加/相减:

- 1 + 4 = 3

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

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

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

3 - 2 cos (x) - 4 cos^{2} (x) = 0

用替代法求解

3 - 2 cos (x) - 4 cos^{2} (x) = 0

令

: cos (x) = u

3 - 2 u - 4 u^{2} = 0

3 - 2 u - 4 u^{2} = 0 : u = - \frac{1 + 13}{4}, u = \frac{13 - 1}{4}

3 - 2 u - 4 u^{2} = 0

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

a = - 4, b = - 2, c = 3

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

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

(- 2)^{2} - 4 (- 4) \cdot 3 = 213

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

使用法则

- (- a) = a

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

使用指数法则:

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

若

n

是偶数

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

= 2^{2} + 4 \cdot 4 \cdot 3

数字相乘:

4 \cdot 4 \cdot 3 = 48

= 2^{2} + 48

2^{2} = 4

= 4 + 48

数字相加:

4 + 48 = 52

= 52

52

质因数分解:

2^{2} \cdot 13

52

52

除以

252 = 26 \cdot 2

= 2 \cdot 26

26

除以

226 = 13 \cdot 2

= 2 \cdot 2 \cdot 13

2, 13

都是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 13

= 2^{2} \cdot 13

= 2^{2} \cdot 13

使用根式运算法则:

n ab = n a n b

= 13 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 213

u_{1, 2} = \frac{- ( - 2 ) \pm 2 13}{2 ( - 4 )}

将解分隔开

u_{1} = \frac{- ( - 2 ) + 2 13}{2 ( - 4 )}, u_{2} = \frac{- ( - 2 ) - 2 13}{2 ( - 4 )}

u = \frac{- ( - 2 ) + 2 13}{2 ( - 4 )} : - \frac{1 + 13}{4}

\frac{- ( - 2 ) + 2 13}{2 ( - 4 )}

去除括号:

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

= \frac{2 + 2 13}{- 2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{2 + 2 13}{- 8}

使用分式法则:

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

= - \frac{2 + 2 13}{8}

消掉

\frac{2 + 2 13}{8} : \frac{1 + 13}{4}

\frac{2 + 2 13}{8}

分解

2 + 213 : 2 (1 + 13)

2 + 213

改写为

= 2 \cdot 1 + 213

因式分解出通项

2

= 2 (1 + 13)

= \frac{2 ( 1 + 13 )}{8}

约分:

2

= \frac{1 + 13}{4}

= - \frac{1 + 13}{4}

u = \frac{- ( - 2 ) - 2 13}{2 ( - 4 )} : \frac{13 - 1}{4}

\frac{- ( - 2 ) - 2 13}{2 ( - 4 )}

去除括号:

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

= \frac{2 - 2 13}{- 2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{2 - 2 13}{- 8}

使用分式法则:

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

2 - 213 = - (213 - 2)

= \frac{2 13 - 2}{8}

分解

213 - 2 : 2 (13 - 1)

213 - 2

改写为

= 213 - 2 \cdot 1

因式分解出通项

2

= 2 (13 - 1)

= \frac{2 ( 13 - 1 )}{8}

约分:

2

= \frac{13 - 1}{4}

二次方程组的解是:

u = - \frac{1 + 13}{4}, u = \frac{13 - 1}{4}

u = cos (x)

代回

cos (x) = - \frac{1 + 13}{4}, cos (x) = \frac{13 - 1}{4}

cos (x) = - \frac{1 + 13}{4}, cos (x) = \frac{13 - 1}{4}

cos (x) = - \frac{1 + 13}{4} :

无解

cos (x) = - \frac{1 + 13}{4}

- 1 \leq cos (x) \leq 1

无解

cos (x) = \frac{13 - 1}{4} : x = arccos (\frac{13 - 1}{4}) + 2 πn, x = 2 π - arccos (\frac{13 - 1}{4}) + 2 πn

cos (x) = \frac{13 - 1}{4}

使用反三角函数性质

cos (x) = \frac{13 - 1}{4}

cos (x) = \frac{13 - 1}{4}

的通解

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

x = arccos (\frac{13 - 1}{4}) + 2 πn, x = 2 π - arccos (\frac{13 - 1}{4}) + 2 πn

x = arccos (\frac{13 - 1}{4}) + 2 πn, x = 2 π - arccos (\frac{13 - 1}{4}) + 2 πn

合并所有解

x = arccos (\frac{13 - 1}{4}) + 2 πn, x = 2 π - arccos (\frac{13 - 1}{4}) + 2 πn

以小数形式表示解

x = 0.86138 \dots + 2 πn, x = 2 π - 0.86138 \dots + 2 πn

作图

流行的例子

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

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

cos^2(2x+pi/6)= 1/2

cos^{2} (2 x + \frac{π}{6}) = \frac{1}{2}

cos(5x)=1,0<= x<= 2pi

cos (5 x) = 1, 0 \leq x \leq 2 π

sec(2θ)=2,0<θ< pi/2

sec (2 θ) = 2, 0 < θ < \frac{π}{2}

2 cos (2 θ) + 1 = 0