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(x)+cos^2(x)= 5/4

解答

sin (x) + cos^{2} (x) = \frac{5}{4}

解答

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

+1

度数

x = 3 0^{\circ} + 36 0^{\circ} n, x = 15 0^{\circ} + 36 0^{\circ} n

求解步骤

sin (x) + cos^{2} (x) = \frac{5}{4}

两边减去

\frac{5}{4}

sin (x) + cos^{2} (x) - \frac{5}{4} = 0

化简

sin (x) + cos^{2} (x) - \frac{5}{4} : \frac{4 sin ( x ) + 4 cos ^{2} ( x ) - 5}{4}

sin (x) + cos^{2} (x) - \frac{5}{4}

将项转换为分式:

sin (x) = \frac{sin ( x ) 4}{4}, cos^{2} (x) = \frac{cos ^{2} ( x ) 4}{4}

= \frac{sin ( x ) \cdot 4}{4} + \frac{cos ^{2} ( x ) \cdot 4}{4} - \frac{5}{4}

因为分母相等，所以合并分式:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{sin ( x ) \cdot 4 + cos ^{2} ( x ) \cdot 4 - 5}{4}

\frac{4 sin ( x ) + 4 cos ^{2} ( x ) - 5}{4} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

4 sin (x) + 4 cos^{2} (x) - 5 = 0

使用三角恒等式改写

- 5 + 4 cos^{2} (x) + 4 sin (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 + 4 (1 - sin^{2} (x)) + 4 sin (x)

化简

- 5 + 4 (1 - sin^{2} (x)) + 4 sin (x) : 4 sin (x) - 4 sin^{2} (x) - 1

- 5 + 4 (1 - sin^{2} (x)) + 4 sin (x)

乘开

4 (1 - sin^{2} (x)) : 4 - 4 sin^{2} (x)

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

使用分配律:

a (b - c) = ab - a c

a = 4, b = 1, c = sin^{2} (x)

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

数字相乘:

4 \cdot 1 = 4

= 4 - 4 sin^{2} (x)

= - 5 + 4 - 4 sin^{2} (x) + 4 sin (x)

数字相加/相减:

- 5 + 4 = - 1

= 4 sin (x) - 4 sin^{2} (x) - 1

= 4 sin (x) - 4 sin^{2} (x) - 1

- 1 + 4 sin (x) - 4 sin^{2} (x) = 0

用替代法求解

- 1 + 4 sin (x) - 4 sin^{2} (x) = 0

令

: sin (x) = u

- 1 + 4 u - 4 u^{2} = 0

- 1 + 4 u - 4 u^{2} = 0 : u = \frac{1}{2}

- 1 + 4 u - 4 u^{2} = 0

改写成标准形式

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

- 4 u^{2} + 4 u - 1 = 0

使用求根公式求解

- 4 u^{2} + 4 u - 1 = 0

二次方程求根公式:

若

a = - 4, b = 4, c = - 1

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

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

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

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

使用法则

- (- a) = a

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

数字相乘:

4 \cdot 4 \cdot 1 = 16

= 4^{2} - 16

4^{2} = 16

= 16 - 16

数字相减:

16 - 16 = 0

= 0

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

u = \frac{- 4}{2 ( - 4 )}

\frac{- 4}{2 ( - 4 )} = \frac{1}{2}

\frac{- 4}{2 ( - 4 )}

去除括号:

(- a) = - a

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

数字相乘:

2 \cdot 4 = 8

= \frac{- 4}{- 8}

使用分式法则:

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

= \frac{4}{8}

约分:

4

= \frac{1}{2}

u = \frac{1}{2}

二次方程组的解是:

u = \frac{1}{2}

u = sin (x)

代回

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

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

sin (x) = \frac{1}{2} : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

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

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{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

合并所有解

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

作图

流行的例子

15sin(x)+6cos(x)-3=0

15 sin (x) + 6 cos (x) - 3 = 0

sqrt(1+cot^2(x))=8

1 + cot^{2} (x) = 8

tan(x)-sqrt(1-2tan^2(x))=0

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

sin(x+pi/6)=(sqrt(2))/2

sin (x + \frac{π}{6}) = \frac{2}{2}

2sin(4x)=sin(2x)

2 sin (4 x) = sin (2 x)