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受欢迎的三角函数 >

cos^2(x)=1-sin(x),0<= x<2pi

解答

cos^{2} (x) = 1 - sin (x), 0 \leq x < 2 π

解答

x = 0, x = π, x = \frac{π}{2}

+1

度数

x = 0^{\circ}, x = 18 0^{\circ}, x = 9 0^{\circ}

求解步骤

cos^{2} (x) = 1 - sin (x), 0 \leq x < 2 π

两边减去

1 - sin (x)

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

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

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

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

用替代法求解

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

令

: sin (x) = u

u - u^{2} = 0

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

u - u^{2} = 0

改写成标准形式

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

- u^{2} + u = 0

使用求根公式求解

- u^{2} + u = 0

二次方程求根公式:

若

a = - 1, b = 1, c = 0

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 1 ) \cdot 0}{2 ( - 1 )}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 1 ) \cdot 0}{2 ( - 1 )}

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

1^{2} - 4 (- 1) \cdot 0

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 1) \cdot 0

使用法则

- (- a) = a

= 1 + 4 \cdot 1 \cdot 0

使用法则

0 \cdot a = 0

= 1 + 0

数字相加:

1 + 0 = 1

= 1

使用法则

1 = 1

= 1

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

将解分隔开

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

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

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

去除括号:

(- a) = - a

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

数字相加/相减:

- 1 + 1 = 0

= \frac{0}{- 2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{0}{- 2}

使用分式法则:

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

= - \frac{0}{2}

使用法则

\frac{0}{a} = 0, a \neq = 0

= - 0

= 0

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

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

去除括号:

(- a) = - a

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

数字相减:

- 1 - 1 = - 2

= \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

二次方程组的解是:

u = 0, u = 1

u = sin (x)

代回

sin (x) = 0, sin (x) = 1

sin (x) = 0, sin (x) = 1

sin (x) = 0, 0 \leq x < 2 π : x = 0, x = π

sin (x) = 0, 0 \leq x < 2 π

sin (x) = 0

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 = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 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, x = π + 2 πn

在

0 \leq x < 2 π

范围内的解

x = 0, x = π

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

sin (x) = 1, 0 \leq x < 2 π

sin (x) = 1

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{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

在

0 \leq x < 2 π

范围内的解

x = \frac{π}{2}

合并所有解

x = 0, x = π, x = \frac{π}{2}

作图

流行的例子

cos(2x)-3sin(x)=2,0<= x<= 360

cos (2 x) - 3 sin (x) = 2, 0^{\circ} \leq x \leq 36 0^{\circ}

2cos^2(x)+3cos(x)=4cos(x)+1

2 cos^{2} (x) + 3 cos (x) = 4 cos (x) + 1

6sin^2(x)+5sin(x)=0

6 sin^{2} (x) + 5 sin (x) = 0

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

sin (x) = - \frac{1}{2}, (0, 2 π)

tan(θ)= 4/3 ,cos(θ)<0

tan (θ) = \frac{4}{3}, cos (θ) < 0