zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的三角函数 >

sqrt(2)sin^2(x)+cos(x)=0

解答

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

解答

x = \frac{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn

+1

度数

x = 13 5^{\circ} + 36 0^{\circ} n, x = 22 5^{\circ} + 36 0^{\circ} n

求解步骤

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

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

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

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

用替代法求解

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

令

: cos (x) = u

u + (1 - u^{2}) 2 = 0

u + (1 - u^{2}) 2 = 0 : u = - \frac{2}{2}, u = 2

u + (1 - u^{2}) 2 = 0

展开

u + (1 - u^{2}) 2 : u + 2 - 2 u^{2}

u + (1 - u^{2}) 2

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

乘开

2 (1 - u^{2}) : 2 - 2 u^{2}

2 (1 - u^{2})

使用分配律:

a (b - c) = ab - a c

a = 2, b = 1, c = u^{2}

= 2 \cdot 1 - 2 u^{2}

= 1 \cdot 2 - 2 u^{2}

乘以:

1 \cdot 2 = 2

= 2 - 2 u^{2}

= u + 2 - 2 u^{2}

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

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

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

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

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

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

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

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 42 (- 2)

使用法则

- (- a) = a

= 1 + 422

422 = 8

422

使用根式运算法则:

a a = a

22 = 2

= 4 \cdot 2

数字相乘:

4 \cdot 2 = 8

= 8

= 1 + 8

数字相加:

1 + 8 = 9

= 9

因式分解数字:

9 = 3^{2}

= 3^{2}

使用根式运算法则:

n a^{n} = a

3^{2} = 3

= 3

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

将解分隔开

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

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

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

去除括号:

(- a) = - a

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

数字相加/相减:

- 1 + 3 = 2

= \frac{2}{- 2 2}

使用分式法则:

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

= - \frac{2}{2 2}

数字相除:

\frac{2}{2} = 1

= - \frac{1}{2}

- \frac{1}{2}

有理化:

- \frac{2}{2}

- \frac{1}{2}

乘以共轭根式

\frac{2}{2}

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

1 \cdot 2 = 2

22 = 2

22

使用根式运算法则:

a a = a

22 = 2

= 2

= - \frac{2}{2}

= - \frac{2}{2}

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

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

去除括号:

(- a) = - a

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

数字相减:

- 1 - 3 = - 4

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

使用分式法则:

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

= \frac{4}{2 2}

数字相除:

\frac{4}{2} = 2

= \frac{2}{2}

使用根式运算法则:

n a = a^{\frac{1}{n}}

2 = 2^{\frac{1}{2}}

= \frac{2}{2 ^{\frac{1}{2}}}

使用指数法则:

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

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

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

数字相减:

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

= 2^{\frac{1}{2}}

使用根式运算法则:

a^{\frac{1}{n}} = n a

2^{\frac{1}{2}} = 2

= 2

二次方程组的解是:

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

u = cos (x)

代回

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

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

cos (x) = - \frac{2}{2} : x = \frac{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn

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

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

的通解

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = \frac{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn

x = \frac{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn

cos (x) = 2 :

无解

cos (x) = 2

- 1 \leq cos (x) \leq 1

无解

合并所有解

x = \frac{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn

作图

流行的例子

sec(3x)= 2/(sqrt(3)),0<= x<= 2pi

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

sec (4 x) - 2 = 0

2cos(θ)sin(θ)=cos(θ)

2 cos (θ) sin (θ) = cos (θ)

- 2 sin (x) = 0

sin(3x)=(sqrt(3))/2

sin (3 x) = \frac{3}{2}