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受欢迎的三角函数 >

tan(2θ)+2cos(θ)=0,0<= θ<= 2pi

解答

tan (2 θ) + 2 cos (θ) = 0, 0 \leq θ \leq 2 π

解答

θ = \frac{π}{2}, θ = \frac{3 π}{2}, θ = \frac{7 π}{6}, θ = \frac{11 π}{6}

+1

度数

θ = 9 0^{\circ}, θ = 27 0^{\circ}, θ = 21 0^{\circ}, θ = 33 0^{\circ}

求解步骤

tan (2 θ) + 2 cos (θ) = 0, 0 \leq θ \leq 2 π

用 sin, cos 表示

tan (2 θ) + 2 cos (θ)

使用基本三角恒等式:

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

= \frac{sin ( 2 θ )}{cos ( 2 θ )} + 2 cos (θ)

化简

\frac{sin ( 2 θ )}{cos ( 2 θ )} + 2 cos (θ) : \frac{sin ( 2 θ ) + 2 cos ( θ ) cos ( 2 θ )}{cos ( 2 θ )}

\frac{sin ( 2 θ )}{cos ( 2 θ )} + 2 cos (θ)

将项转换为分式:

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

= \frac{sin ( 2 θ )}{cos ( 2 θ )} + \frac{2 cos ( θ ) cos ( 2 θ )}{cos ( 2 θ )}

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 ( 2 θ ) + 2 cos ( θ ) cos ( 2 θ )}{cos ( 2 θ )}

= \frac{sin ( 2 θ ) + 2 cos ( θ ) cos ( 2 θ )}{cos ( 2 θ )}

\frac{sin ( 2 θ ) + 2 cos ( 2 θ ) cos ( θ )}{cos ( 2 θ )} = 0

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

sin (2 θ) + 2 cos (2 θ) cos (θ) = 0

使用三角恒等式改写

sin (2 θ) + 2 cos (2 θ) cos (θ)

使用倍角公式:

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

= 2 sin (θ) cos (θ) + 2 cos (2 θ) cos (θ)

2 cos (2 θ) cos (θ) + 2 cos (θ) sin (θ) = 0

分解

2 cos (2 θ) cos (θ) + 2 cos (θ) sin (θ) : 2 cos (θ) (cos (2 θ) + sin (θ))

2 cos (2 θ) cos (θ) + 2 cos (θ) sin (θ)

改写为

= 2 cos (θ) cos (2 θ) + 2 cos (θ) sin (θ)

因式分解出通项

2 cos (θ)

= 2 cos (θ) (cos (2 θ) + sin (θ))

2 cos (θ) (cos (2 θ) + sin (θ)) = 0

分别求解每个部分

cos (θ) = 0 or cos (2 θ) + sin (θ) = 0

cos (θ) = 0, 0 \leq θ \leq 2 π : θ = \frac{π}{2}, θ = \frac{3 π}{2}

cos (θ) = 0, 0 \leq θ \leq 2 π

cos (θ) = 0

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}

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

在

0 \leq θ \leq 2 π

范围内的解

θ = \frac{π}{2}, θ = \frac{3 π}{2}

cos (2 θ) + sin (θ) = 0, 0 \leq θ \leq 2 π : θ = \frac{7 π}{6}, θ = \frac{11 π}{6}, θ = \frac{π}{2}

cos (2 θ) + sin (θ) = 0, 0 \leq θ \leq 2 π

使用三角恒等式改写

cos (2 θ) + sin (θ)

使用倍角公式:

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

= 1 - 2 sin^{2} (θ) + sin (θ)

1 + sin (θ) - 2 sin^{2} (θ) = 0

用替代法求解

1 + sin (θ) - 2 sin^{2} (θ) = 0

令

: sin (θ) = u

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

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

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

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

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

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

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

1^{2} - 4 (- 2) \cdot 1 = 3

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

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 2) \cdot 1

使用法则

- (- a) = a

= 1 + 4 \cdot 2 \cdot 1

数字相乘:

4 \cdot 2 \cdot 1 = 8

= 1 + 8

数字相加:

1 + 8 = 9

= 9

因式分解数字:

9 = 3^{2}

= 3^{2}

使用根式运算法则:

n a^{n} = a

3^{2} = 3

= 3

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

将解分隔开

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

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

\frac{- 1 + 3}{2 ( - 2 )}

去除括号:

(- a) = - a

= \frac{- 1 + 3}{- 2 \cdot 2}

数字相加/相减:

- 1 + 3 = 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}

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

\frac{- 1 - 3}{2 ( - 2 )}

去除括号:

(- a) = - a

= \frac{- 1 - 3}{- 2 \cdot 2}

数字相减:

- 1 - 3 = - 4

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

数字相乘:

2 \cdot 2 = 4

= \frac{- 4}{- 4}

使用分式法则:

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

= \frac{4}{4}

使用法则

\frac{a}{a} = 1

= 1

二次方程组的解是:

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

u = sin (θ)

代回

sin (θ) = - \frac{1}{2}, sin (θ) = 1

sin (θ) = - \frac{1}{2}, sin (θ) = 1

sin (θ) = - \frac{1}{2}, 0 \leq θ \leq 2 π : θ = \frac{7 π}{6}, θ = \frac{11 π}{6}

sin (θ) = - \frac{1}{2}, 0 \leq θ \leq 2 π

sin (θ) = - \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}

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

在

0 \leq θ \leq 2 π

范围内的解

θ = \frac{7 π}{6}, θ = \frac{11 π}{6}

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

sin (θ) = 1, 0 \leq θ \leq 2 π

sin (θ) = 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}

θ = \frac{π}{2} + 2 πn

θ = \frac{π}{2} + 2 πn

在

0 \leq θ \leq 2 π

范围内的解

θ = \frac{π}{2}

合并所有解

θ = \frac{7 π}{6}, θ = \frac{11 π}{6}, θ = \frac{π}{2}

合并所有解

θ = \frac{π}{2}, θ = \frac{3 π}{2}, θ = \frac{7 π}{6}, θ = \frac{11 π}{6}

作图

流行的例子

cos(2x)+10sin^2(x)=5

cos (2 x) + 10 sin^{2} (x) = 5

2 = 3 + 2 sin (θ)

-2sin(θ)+sin(2θ)=0

- 2 sin (θ) + sin (2 θ) = 0

-2sin^2(x)-3sin(x)+2=0

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

tan (β) = \frac{1}{2}