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受欢迎的三角函数 >

sin^2(x)-5cos(x)=3

解答

sin^{2} (x) - 5 cos (x) = 3

解答

x = 2.02466 \dots + 2 πn, x = - 2.02466 \dots + 2 πn

+1

度数

x = 116.00484 \dots^{\circ} + 36 0^{\circ} n, x = - 116.00484 \dots^{\circ} + 36 0^{\circ} n

求解步骤

sin^{2} (x) - 5 cos (x) = 3

两边减去

3

sin^{2} (x) - 5 cos (x) - 3 = 0

使用三角恒等式改写

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

= - 3 + 1 - cos^{2} (x) - 5 cos (x)

化简

= - cos^{2} (x) - 5 cos (x) - 2

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

用替代法求解

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

令

: cos (x) = u

- 2 - u^{2} - 5 u = 0

- 2 - u^{2} - 5 u = 0 : u = - \frac{5 + 17}{2}, u = - \frac{5 - 17}{2}

- 2 - u^{2} - 5 u = 0

改写成标准形式

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

- u^{2} - 5 u - 2 = 0

使用求根公式求解

- u^{2} - 5 u - 2 = 0

二次方程求根公式:

若

a = - 1, b = - 5, c = - 2

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 ( - 1 ) ( - 2 )}{2 ( - 1 )}

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 ( - 1 ) ( - 2 )}{2 ( - 1 )}

(- 5)^{2} - 4 (- 1) (- 2) = 17

(- 5)^{2} - 4 (- 1) (- 2)

使用法则

- (- a) = a

= (- 5)^{2} - 4 \cdot 1 \cdot 2

使用指数法则:

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

若

n

是偶数

(- 5)^{2} = 5^{2}

= 5^{2} - 4 \cdot 1 \cdot 2

数字相乘:

4 \cdot 1 \cdot 2 = 8

= 5^{2} - 8

5^{2} = 25

= 25 - 8

数字相减:

25 - 8 = 17

= 17

u_{1, 2} = \frac{- ( - 5 ) \pm 17}{2 ( - 1 )}

将解分隔开

u_{1} = \frac{- ( - 5 ) + 17}{2 ( - 1 )}, u_{2} = \frac{- ( - 5 ) - 17}{2 ( - 1 )}

u = \frac{- ( - 5 ) + 17}{2 ( - 1 )} : - \frac{5 + 17}{2}

\frac{- ( - 5 ) + 17}{2 ( - 1 )}

去除括号:

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

= \frac{5 + 17}{- 2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{5 + 17}{- 2}

使用分式法则:

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

= - \frac{5 + 17}{2}

u = \frac{- ( - 5 ) - 17}{2 ( - 1 )} : - \frac{5 - 17}{2}

\frac{- ( - 5 ) - 17}{2 ( - 1 )}

去除括号:

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

= \frac{5 - 17}{- 2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{5 - 17}{- 2}

使用分式法则:

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

= - \frac{5 - 17}{2}

二次方程组的解是:

u = - \frac{5 + 17}{2}, u = - \frac{5 - 17}{2}

u = cos (x)

代回

cos (x) = - \frac{5 + 17}{2}, cos (x) = - \frac{5 - 17}{2}

cos (x) = - \frac{5 + 17}{2}, cos (x) = - \frac{5 - 17}{2}

cos (x) = - \frac{5 + 17}{2} :

无解

cos (x) = - \frac{5 + 17}{2}

- 1 \leq cos (x) \leq 1

无解

cos (x) = - \frac{5 - 17}{2} : x = arccos (- \frac{5 - 17}{2}) + 2 πn, x = - arccos (- \frac{5 - 17}{2}) + 2 πn

cos (x) = - \frac{5 - 17}{2}

使用反三角函数性质

cos (x) = - \frac{5 - 17}{2}

cos (x) = - \frac{5 - 17}{2}

的通解

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

x = arccos (- \frac{5 - 17}{2}) + 2 πn, x = - arccos (- \frac{5 - 17}{2}) + 2 πn

x = arccos (- \frac{5 - 17}{2}) + 2 πn, x = - arccos (- \frac{5 - 17}{2}) + 2 πn

合并所有解

x = arccos (- \frac{5 - 17}{2}) + 2 πn, x = - arccos (- \frac{5 - 17}{2}) + 2 πn

以小数形式表示解

x = 2.02466 \dots + 2 πn, x = - 2.02466 \dots + 2 πn

作图

流行的例子

cos (x) = - \frac{9}{41}

cos(θ)=-0.42,0<= θ<360

cos (θ) = - 0.42, 0^{\circ} \leq θ < 36 0^{\circ}

sin(2x)= 1/2 ,0<x<2pi

sin (2 x) = \frac{1}{2}, 0 < x < 2 π

4sin^2(θ)+2cos(θ)=3

4 sin^{2} (θ) + 2 cos (θ) = 3

1715/70 =tan(45+θ/2)

\frac{1715}{70} = tan (4 5^{\circ} + \frac{θ}{2})