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)=2-2cos(x),0<= x<= 2pi

解答

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

解答

x = 0, x = 2 π

+1

度数

x = 0^{\circ}, x = 36 0^{\circ}

求解步骤

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

两边减去

2 - 2 cos (x)

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

使用三角恒等式改写

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

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

化简

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

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

用替代法求解

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

令

: cos (x) = u

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

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

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

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

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

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

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

2^{2} - 4 (- 1) (- 1) = 0

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

使用法则

- (- a) = a

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

数字相乘:

4 \cdot 1 \cdot 1 = 4

= 2^{2} - 4

2^{2} = 4

= 4 - 4

数字相减:

4 - 4 = 0

= 0

u_{1, 2} = \frac{- 2 \pm 0}{2 ( - 1 )}

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

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

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

去除括号:

(- a) = - a

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

数字相乘:

2 \cdot 1 = 2

= \frac{- 2}{- 2}

使用分式法则:

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

= \frac{2}{2}

使用法则

\frac{a}{a} = 1

= 1

u = 1

二次方程组的解是:

u = 1

u = cos (x)

代回

cos (x) = 1

cos (x) = 1

cos (x) = 1, 0 \leq x \leq 2 π : x = 0, x = 2 π

cos (x) = 1, 0 \leq x \leq 2 π

cos (x) = 1

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 = 0 + 2 πn

x = 0 + 2 πn

解

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn

在

0 \leq x \leq 2 π

范围内的解

x = 0, x = 2 π

合并所有解

x = 0, x = 2 π

作图

流行的例子

tan(θ)=(sqrt(7))/(21)

tan (θ) = \frac{7}{21}

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

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

sin (θ - 4 0^{\circ}) = 1

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

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

cos(3x)=cos(6x)

cos (3 x) = cos (6 x)