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 >

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

解答

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

解答

x = 0.66623 \dots + 2 πn, x = π - 0.66623 \dots + 2 πn

+1

度数

x = 38.17270 \dots^{\circ} + 36 0^{\circ} n, x = 141.82729 \dots^{\circ} + 36 0^{\circ} n

求解步骤

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

两边减去

1

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

使用三角恒等式改写

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

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

化简

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

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

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

化简

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

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

对同类项分组

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

同类项相加:

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

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

数字相加/相减:

- 1 + 2 = 1

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

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

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

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

用替代法求解

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

令

: sin (x) = u

1 - u - u^{2} = 0

1 - u - u^{2} = 0 : u = - \frac{1 + 5}{2}, u = \frac{5 - 1}{2}

1 - u - u^{2} = 0

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

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

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

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

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

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

使用法则

- (- a) = a

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

4 \cdot 1 \cdot 1

数字相乘:

4 \cdot 1 \cdot 1 = 4

= 4

= 1 + 4

数字相加:

1 + 4 = 5

= 5

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

将解分隔开

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

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

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

去除括号:

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

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

数字相乘:

2 \cdot 1 = 2

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

使用分式法则:

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

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

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

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

去除括号:

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

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

数字相乘:

2 \cdot 1 = 2

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

使用分式法则:

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

1 - 5 = - (5 - 1)

= \frac{5 - 1}{2}

二次方程组的解是:

u = - \frac{1 + 5}{2}, u = \frac{5 - 1}{2}

u = sin (x)

代回

sin (x) = - \frac{1 + 5}{2}, sin (x) = \frac{5 - 1}{2}

sin (x) = - \frac{1 + 5}{2}, sin (x) = \frac{5 - 1}{2}

sin (x) = - \frac{1 + 5}{2} :

无解

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

- 1 \leq sin (x) \leq 1

无解

sin (x) = \frac{5 - 1}{2} : x = arcsin (\frac{5 - 1}{2}) + 2 πn, x = π - arcsin (\frac{5 - 1}{2}) + 2 πn

sin (x) = \frac{5 - 1}{2}

使用反三角函数性质

sin (x) = \frac{5 - 1}{2}

sin (x) = \frac{5 - 1}{2}

的通解

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

x = arcsin (\frac{5 - 1}{2}) + 2 πn, x = π - arcsin (\frac{5 - 1}{2}) + 2 πn

x = arcsin (\frac{5 - 1}{2}) + 2 πn, x = π - arcsin (\frac{5 - 1}{2}) + 2 πn

合并所有解

x = arcsin (\frac{5 - 1}{2}) + 2 πn, x = π - arcsin (\frac{5 - 1}{2}) + 2 πn

以小数形式表示解

x = 0.66623 \dots + 2 πn, x = π - 0.66623 \dots + 2 πn

作图

流行的例子

(sin^3(x))/(2+2(sin(x))^2)=1

\frac{sin ^{3} ( x )}{2 + 2 ( sin ( x ) ) ^{2}} = 1

solvefor x,cot^2(x)= 1/3

so l v e f or x, cot^{2} (x) = \frac{1}{3}

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

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

tan^5(x)tan^2(x)=1

tan^{5} (x) tan^{2} (x) = 1

(sin^2(a)+1)/((1+tan^2(a)))=1

\frac{sin ^{2} ( a ) + 1}{( 1 + tan ^{2} ( a ) )} = 1