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受欢迎的三角函数 >

2sin(x+pi/2)+3tan(pi-x)=0

解答

2 sin (x + \frac{π}{2}) + 3 tan (π - x) = 0

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

2 sin (x + \frac{π}{2}) + 3 tan (π - x) = 0

使用三角恒等式改写

2 sin (x + \frac{π}{2}) + 3 tan (π - x) = 0

使用三角恒等式改写

sin (x + \frac{π}{2})

使用角和恒等式:

sin (s + t) = sin (s) cos (t) + cos (s) sin (t)

= sin (x) cos (\frac{π}{2}) + cos (x) sin (\frac{π}{2})

化简

sin (x) cos (\frac{π}{2}) + cos (x) sin (\frac{π}{2}) : cos (x)

sin (x) cos (\frac{π}{2}) + cos (x) sin (\frac{π}{2})

sin (x) cos (\frac{π}{2}) = 0

sin (x) cos (\frac{π}{2})

化简

cos (\frac{π}{2}) : 0

cos (\frac{π}{2})

使用以下普通恒等式:

cos (\frac{π}{2}) = 0

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}

= 0

= 0 \cdot sin (x)

使用法则

0 \cdot a = 0

= 0

cos (x) sin (\frac{π}{2}) = cos (x)

cos (x) sin (\frac{π}{2})

化简

sin (\frac{π}{2}) : 1

sin (\frac{π}{2})

使用以下普通恒等式:

sin (\frac{π}{2}) = 1

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}

= 1

= 1 \cdot cos (x)

乘以:

cos (x) \cdot 1 = cos (x)

= cos (x)

= 0 + cos (x)

0 + cos (x) = cos (x)

= cos (x)

= cos (x)

使用基本三角恒等式:

tan (x) = \frac{sin ( x )}{cos ( x )}

= \frac{sin ( π - x )}{cos ( π - x )}

使用角差恒等式:

sin (s - t) = sin (s) cos (t) - cos (s) sin (t)

= \frac{sin ( π ) cos ( x ) - cos ( π ) sin ( x )}{cos ( π - x )}

使用角差恒等式:

cos (s - t) = cos (s) cos (t) + sin (s) sin (t)

= \frac{sin ( π ) cos ( x ) - cos ( π ) sin ( x )}{cos ( π ) cos ( x ) + sin ( π ) sin ( x )}

化简

\frac{sin ( π ) cos ( x ) - cos ( π ) sin ( x )}{cos ( π ) cos ( x ) + sin ( π ) sin ( x )} : - \frac{sin ( x )}{cos ( x )}

\frac{sin ( π ) cos ( x ) - cos ( π ) sin ( x )}{cos ( π ) cos ( x ) + sin ( π ) sin ( x )}

sin (π) cos (x) - cos (π) sin (x) = sin (x)

sin (π) cos (x) - cos (π) sin (x)

sin (π) cos (x) = 0

sin (π) cos (x)

化简

sin (π) : 0

sin (π)

使用以下普通恒等式:

sin (π) = 0

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}

= 0

= 0 \cdot cos (x)

使用法则

0 \cdot a = 0

= 0

cos (π) sin (x) = - sin (x)

cos (π) sin (x)

化简

cos (π) : - 1

cos (π)

使用以下普通恒等式:

cos (π) = (- 1)

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}

= - 1

= - 1 \cdot sin (x)

乘以:

1 \cdot sin (x) = sin (x)

= - sin (x)

= 0 - (- sin (x))

整理后得

= sin (x)

= \frac{sin ( x )}{cos ( π ) cos ( x ) + sin ( π ) sin ( x )}

cos (π) cos (x) + sin (π) sin (x) = - cos (x)

cos (π) cos (x) + sin (π) sin (x)

cos (π) cos (x) = - cos (x)

cos (π) cos (x)

化简

cos (π) : - 1

cos (π)

使用以下普通恒等式:

cos (π) = (- 1)

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}

= - 1

= - 1 \cdot cos (x)

乘以:

1 \cdot cos (x) = cos (x)

= - cos (x)

= - cos (x) + sin (π) sin (x)

sin (π) sin (x) = 0

sin (π) sin (x)

化简

sin (π) : 0

sin (π)

使用以下普通恒等式:

sin (π) = 0

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}

= 0

= 0 \cdot sin (x)

使用法则

0 \cdot a = 0

= 0

= - cos (x) + 0

- cos (x) + 0 = - cos (x)

= - cos (x)

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

使用分式法则:

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

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

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

2 cos (x) + 3 (- \frac{sin ( x )}{cos ( x )}) = 0

化简

2 cos (x) + 3 (- \frac{sin ( x )}{cos ( x )}) : 2 cos (x) - \frac{3 sin ( x )}{cos ( x )}

2 cos (x) + 3 (- \frac{sin ( x )}{cos ( x )})

去除括号:

(- a) = - a

= 2 cos (x) - 3 \cdot \frac{sin ( x )}{cos ( x )}

乘

3 \cdot \frac{sin ( x )}{cos ( x )} : \frac{3 sin ( x )}{cos ( x )}

3 \cdot \frac{sin ( x )}{cos ( x )}

分式相乘:

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

= \frac{sin ( x ) \cdot 3}{cos ( x )}

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

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

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

化简

2 cos (x) - \frac{3 sin ( x )}{cos ( x )} : \frac{2 cos ^{2} ( x ) - 3 sin ( x )}{cos ( x )}

2 cos (x) - \frac{3 sin ( x )}{cos ( x )}

将项转换为分式:

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

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

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{2 cos ( x ) cos ( x ) - 3 sin ( x )}{cos ( x )}

2 cos (x) cos (x) - 3 sin (x) = 2 cos^{2} (x) - 3 sin (x)

2 cos (x) cos (x) - 3 sin (x)

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

2 cos (x) cos (x)

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

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

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

数字相加:

1 + 1 = 2

= 2 cos^{2} (x)

= 2 cos^{2} (x) - 3 sin (x)

= \frac{2 cos ^{2} ( x ) - 3 sin ( x )}{cos ( x )}

\frac{2 cos ^{2} ( x ) - 3 sin ( x )}{cos ( x )} = 0

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

2 cos^{2} (x) - 3 sin (x) = 0

使用三角恒等式改写

2 cos^{2} (x) - 3 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)

= 2 (1 - sin^{2} (x)) - 3 sin (x)

(1 - sin^{2} (x)) \cdot 2 - 3 sin (x) = 0

用替代法求解

(1 - sin^{2} (x)) \cdot 2 - 3 sin (x) = 0

令

: sin (x) = u

(1 - u^{2}) \cdot 2 - 3 u = 0

(1 - u^{2}) \cdot 2 - 3 u = 0 : u = - 2, u = \frac{1}{2}

(1 - u^{2}) \cdot 2 - 3 u = 0

展开

(1 - u^{2}) \cdot 2 - 3 u : 2 - 2 u^{2} - 3 u

(1 - u^{2}) \cdot 2 - 3 u

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

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}

数字相乘:

2 \cdot 1 = 2

= 2 - 2 u^{2}

= 2 - 2 u^{2} - 3 u

2 - 2 u^{2} - 3 u = 0

改写成标准形式

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

- 2 u^{2} - 3 u + 2 = 0

使用求根公式求解

- 2 u^{2} - 3 u + 2 = 0

二次方程求根公式:

若

a = - 2, b = - 3, c = 2

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 ( - 2 ) \cdot 2}{2 ( - 2 )}

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 ( - 2 ) \cdot 2}{2 ( - 2 )}

(- 3)^{2} - 4 (- 2) \cdot 2 = 5

(- 3)^{2} - 4 (- 2) \cdot 2

使用法则

- (- a) = a

= (- 3)^{2} + 4 \cdot 2 \cdot 2

使用指数法则:

(- a)^{n} = a^{n},

若

n

是偶数

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

= 3^{2} + 4 \cdot 2 \cdot 2

数字相乘:

4 \cdot 2 \cdot 2 = 16

= 3^{2} + 16

3^{2} = 9

= 9 + 16

数字相加:

9 + 16 = 25

= 25

因式分解数字:

25 = 5^{2}

= 5^{2}

使用根式运算法则:

n a^{n} = a

5^{2} = 5

= 5

u_{1, 2} = \frac{- ( - 3 ) \pm 5}{2 ( - 2 )}

将解分隔开

u_{1} = \frac{- ( - 3 ) + 5}{2 ( - 2 )}, u_{2} = \frac{- ( - 3 ) - 5}{2 ( - 2 )}

u = \frac{- ( - 3 ) + 5}{2 ( - 2 )} : - 2

\frac{- ( - 3 ) + 5}{2 ( - 2 )}

去除括号:

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

= \frac{3 + 5}{- 2 \cdot 2}

数字相加:

3 + 5 = 8

= \frac{8}{- 2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{8}{- 4}

使用分式法则:

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

= - \frac{8}{4}

数字相除:

\frac{8}{4} = 2

= - 2

u = \frac{- ( - 3 ) - 5}{2 ( - 2 )} : \frac{1}{2}

\frac{- ( - 3 ) - 5}{2 ( - 2 )}

去除括号:

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

= \frac{3 - 5}{- 2 \cdot 2}

数字相减:

3 - 5 = - 2

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

数字相乘:

2 \cdot 2 = 4

= \frac{- 2}{- 4}

使用分式法则:

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

= \frac{2}{4}

约分:

2

= \frac{1}{2}

二次方程组的解是:

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

u = sin (x)

代回

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

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

sin (x) = - 2 :

无解

sin (x) = - 2

- 1 \leq sin (x) \leq 1

无解

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

sin (x) = 0.27

sec(x/2)=cos(x/2)

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

sec^2(x)-2tan^2(x)=0

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

tan(5x)+sqrt(3)=0

tan (5 x) + 3 = 0

2tan(x)sin(x)-7sin(x)=0

2 tan (x) sin (x) - 7 sin (x) = 0