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受欢迎的三角函数 >

6cos^2(x)-sin(x)-5=0,0<= x<= 2pi

解答

6 cos^{2} (x) - sin (x) - 5 = 0, 0 \leq x \leq 2 π

解答

x = \frac{7 π}{6}, x = \frac{11 π}{6}, x = 0.33983 \dots, x = π - 0.33983 \dots

+1

度数

x = 21 0^{\circ}, x = 33 0^{\circ}, x = 19.47122 \dots^{\circ}, x = 160.52877 \dots^{\circ}

求解步骤

6 cos^{2} (x) - sin (x) - 5 = 0, 0 \leq x \leq 2 π

使用三角恒等式改写

- 5 - sin (x) + 6 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) + 6 (1 - sin^{2} (x))

化简

- 5 - sin (x) + 6 (1 - sin^{2} (x)) : - 6 sin^{2} (x) - sin (x) + 1

- 5 - sin (x) + 6 (1 - sin^{2} (x))

乘开

6 (1 - sin^{2} (x)) : 6 - 6 sin^{2} (x)

6 (1 - sin^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = 6, b = 1, c = sin^{2} (x)

= 6 \cdot 1 - 6 sin^{2} (x)

数字相乘:

6 \cdot 1 = 6

= 6 - 6 sin^{2} (x)

= - 5 - sin (x) + 6 - 6 sin^{2} (x)

化简

- 5 - sin (x) + 6 - 6 sin^{2} (x) : - 6 sin^{2} (x) - sin (x) + 1

- 5 - sin (x) + 6 - 6 sin^{2} (x)

对同类项分组

= - sin (x) - 6 sin^{2} (x) - 5 + 6

数字相加/相减:

- 5 + 6 = 1

= - 6 sin^{2} (x) - sin (x) + 1

= - 6 sin^{2} (x) - sin (x) + 1

= - 6 sin^{2} (x) - sin (x) + 1

1 - sin (x) - 6 sin^{2} (x) = 0

用替代法求解

1 - sin (x) - 6 sin^{2} (x) = 0

令

: sin (x) = u

1 - u - 6 u^{2} = 0

1 - u - 6 u^{2} = 0 : u = - \frac{1}{2}, u = \frac{1}{3}

1 - u - 6 u^{2} = 0

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

a = - 6, b = - 1, c = 1

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 ( - 6 ) \cdot 1}{2 ( - 6 )}

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 ( - 6 ) \cdot 1}{2 ( - 6 )}

(- 1)^{2} - 4 (- 6) \cdot 1 = 5

(- 1)^{2} - 4 (- 6) \cdot 1

使用法则

- (- a) = a

= (- 1)^{2} + 4 \cdot 6 \cdot 1

(- 1)^{2} = 1

(- 1)^{2}

使用指数法则:

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

若

n

是偶数

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

= 1^{2}

使用法则

1^{a} = 1

= 1

4 \cdot 6 \cdot 1 = 24

4 \cdot 6 \cdot 1

数字相乘:

4 \cdot 6 \cdot 1 = 24

= 24

= 1 + 24

数字相加:

1 + 24 = 25

= 25

因式分解数字:

25 = 5^{2}

= 5^{2}

使用根式运算法则:

n a^{n} = a

5^{2} = 5

= 5

u_{1, 2} = \frac{- ( - 1 ) \pm 5}{2 ( - 6 )}

将解分隔开

u_{1} = \frac{- ( - 1 ) + 5}{2 ( - 6 )}, u_{2} = \frac{- ( - 1 ) - 5}{2 ( - 6 )}

u = \frac{- ( - 1 ) + 5}{2 ( - 6 )} : - \frac{1}{2}

\frac{- ( - 1 ) + 5}{2 ( - 6 )}

去除括号:

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

= \frac{1 + 5}{- 2 \cdot 6}

数字相加:

1 + 5 = 6

= \frac{6}{- 2 \cdot 6}

数字相乘:

2 \cdot 6 = 12

= \frac{6}{- 12}

使用分式法则:

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

= - \frac{6}{12}

约分:

6

= - \frac{1}{2}

u = \frac{- ( - 1 ) - 5}{2 ( - 6 )} : \frac{1}{3}

\frac{- ( - 1 ) - 5}{2 ( - 6 )}

去除括号:

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

= \frac{1 - 5}{- 2 \cdot 6}

数字相减:

1 - 5 = - 4

= \frac{- 4}{- 2 \cdot 6}

数字相乘:

2 \cdot 6 = 12

= \frac{- 4}{- 12}

使用分式法则:

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

= \frac{4}{12}

约分:

4

= \frac{1}{3}

二次方程组的解是:

u = - \frac{1}{2}, u = \frac{1}{3}

u = sin (x)

代回

sin (x) = - \frac{1}{2}, sin (x) = \frac{1}{3}

sin (x) = - \frac{1}{2}, sin (x) = \frac{1}{3}

sin (x) = - \frac{1}{2}, 0 \leq x \leq 2 π : x = \frac{7 π}{6}, x = \frac{11 π}{6}

sin (x) = - \frac{1}{2}, 0 \leq x \leq 2 π

sin (x) = - \frac{1}{2}

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}

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

在

0 \leq x \leq 2 π

范围内的解

x = \frac{7 π}{6}, x = \frac{11 π}{6}

sin (x) = \frac{1}{3}, 0 \leq x \leq 2 π : x = arcsin (\frac{1}{3}), x = π - arcsin (\frac{1}{3})

sin (x) = \frac{1}{3}, 0 \leq x \leq 2 π

使用反三角函数性质

sin (x) = \frac{1}{3}

sin (x) = \frac{1}{3}

的通解

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

x = arcsin (\frac{1}{3}) + 2 πn, x = π - arcsin (\frac{1}{3}) + 2 πn

x = arcsin (\frac{1}{3}) + 2 πn, x = π - arcsin (\frac{1}{3}) + 2 πn

在

0 \leq x \leq 2 π

范围内的解

x = arcsin (\frac{1}{3}), x = π - arcsin (\frac{1}{3})

合并所有解

x = \frac{7 π}{6}, x = \frac{11 π}{6}, x = arcsin (\frac{1}{3}), x = π - arcsin (\frac{1}{3})

以小数形式表示解

x = \frac{7 π}{6}, x = \frac{11 π}{6}, x = 0.33983 \dots, x = π - 0.33983 \dots

作图

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