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受欢迎的三角函数 >

4sin(θ)=sqrt(3)sec(θ),0<= θ<180

解答

4 sin (θ) = 3 sec (θ), 0 \leq θ < 18 0^{\circ}

解答

θ = 3 0^{\circ}, θ = 6 0^{\circ}

+1

弧度

θ = \frac{π}{6}, θ = \frac{π}{3}

求解步骤

4 sin (θ) = 3 sec (θ), 0 \leq θ < 18 0^{\circ}

两边减去

3 sec (θ)

4 sin (θ) - 3 sec (θ) = 0

用 sin, cos 表示

4 sin (θ) - sec (θ) 3

使用基本三角恒等式:

sec (x) = \frac{1}{cos ( x )}

= 4 sin (θ) - \frac{1}{cos ( θ )} 3

化简

4 sin (θ) - \frac{1}{cos ( θ )} 3 : \frac{4 sin ( θ ) cos ( θ ) - 3}{cos ( θ )}

4 sin (θ) - \frac{1}{cos ( θ )} 3

\frac{1}{cos ( θ )} 3 = \frac{3}{cos ( θ )}

\frac{1}{cos ( θ )} 3

分式相乘:

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

= \frac{1 \cdot 3}{cos ( θ )}

乘以:

1 \cdot 3 = 3

= \frac{3}{cos ( θ )}

= 4 sin (θ) - \frac{3}{cos ( θ )}

将项转换为分式:

4 sin (θ) = \frac{4 sin ( θ ) cos ( θ )}{cos ( θ )}

= \frac{4 sin ( θ ) cos ( θ )}{cos ( θ )} - \frac{3}{cos ( 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{4 sin ( θ ) cos ( θ ) - 3}{cos ( θ )}

= \frac{4 sin ( θ ) cos ( θ ) - 3}{cos ( θ )}

\frac{- 3 + 4 cos ( θ ) sin ( θ )}{cos ( θ )} = 0

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

- 3 + 4 cos (θ) sin (θ) = 0

使用三角恒等式改写

- 3 + 4 cos (θ) sin (θ)

使用倍角公式:

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

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

= - 3 + 4 \cdot \frac{sin ( 2 θ )}{2}

- 3 + 4 \cdot \frac{sin ( 2 θ )}{2} = 0

4 \cdot \frac{sin ( 2 θ )}{2} = 2 sin (2 θ)

4 \cdot \frac{sin ( 2 θ )}{2}

分式相乘:

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

= \frac{sin ( 2 θ ) \cdot 4}{2}

数字相除:

\frac{4}{2} = 2

= 2 sin (2 θ)

- 3 + 2 sin (2 θ) = 0

将

3

到右边

- 3 + 2 sin (2 θ) = 0

两边加上

3

- 3 + 2 sin (2 θ) + 3 = 0 + 3

化简

2 sin (2 θ) = 3

2 sin (2 θ) = 3

两边除以

2

2 sin (2 θ) = 3

两边除以

2

\frac{2 sin ( 2 θ )}{2} = \frac{3}{2}

化简

sin (2 θ) = \frac{3}{2}

sin (2 θ) = \frac{3}{2}

sin (2 θ) = \frac{3}{2}

的通解

sin (x)

周期表（周期为

36 0^{\circ} n

"）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

2 θ = 6 0^{\circ} + 36 0^{\circ} n, 2 θ = 12 0^{\circ} + 36 0^{\circ} n

2 θ = 6 0^{\circ} + 36 0^{\circ} n, 2 θ = 12 0^{\circ} + 36 0^{\circ} n

解

2 θ = 6 0^{\circ} + 36 0^{\circ} n : θ = 3 0^{\circ} + 18 0^{\circ} n

2 θ = 6 0^{\circ} + 36 0^{\circ} n

两边除以

2

2 θ = 6 0^{\circ} + 36 0^{\circ} n

两边除以

2

\frac{2 θ}{2} = \frac{6 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

化简

\frac{2 θ}{2} = \frac{6 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

化简

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

数字相除:

\frac{2}{2} = 1

= θ

化简

\frac{6 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2} : 3 0^{\circ} + 18 0^{\circ} n

\frac{6 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

\frac{6 0 ^{\circ}}{2} = 3 0^{\circ}

\frac{6 0 ^{\circ}}{2}

使用分式法则:

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

= \frac{18 0 ^{\circ}}{3 \cdot 2}

数字相乘:

3 \cdot 2 = 6

= 3 0^{\circ}

\frac{36 0 ^{\circ} n}{2} = 18 0^{\circ} n

\frac{36 0 ^{\circ} n}{2}

数字相除:

\frac{2}{2} = 1

= 18 0^{\circ} n

= 3 0^{\circ} + 18 0^{\circ} n

θ = 3 0^{\circ} + 18 0^{\circ} n

θ = 3 0^{\circ} + 18 0^{\circ} n

θ = 3 0^{\circ} + 18 0^{\circ} n

解

2 θ = 12 0^{\circ} + 36 0^{\circ} n : θ = 6 0^{\circ} + 18 0^{\circ} n

2 θ = 12 0^{\circ} + 36 0^{\circ} n

两边除以

2

2 θ = 12 0^{\circ} + 36 0^{\circ} n

两边除以

2

\frac{2 θ}{2} = \frac{12 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

化简

\frac{2 θ}{2} = \frac{12 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

化简

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

数字相除:

\frac{2}{2} = 1

= θ

化简

\frac{12 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2} : 6 0^{\circ} + 18 0^{\circ} n

\frac{12 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

\frac{12 0 ^{\circ}}{2} = 6 0^{\circ}

\frac{12 0 ^{\circ}}{2}

使用分式法则:

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

= \frac{36 0 ^{\circ}}{3 \cdot 2}

数字相乘:

3 \cdot 2 = 6

= 6 0^{\circ}

约分:

2

= 6 0^{\circ}

\frac{36 0 ^{\circ} n}{2} = 18 0^{\circ} n

\frac{36 0 ^{\circ} n}{2}

数字相除:

\frac{2}{2} = 1

= 18 0^{\circ} n

= 6 0^{\circ} + 18 0^{\circ} n

θ = 6 0^{\circ} + 18 0^{\circ} n

θ = 6 0^{\circ} + 18 0^{\circ} n

θ = 6 0^{\circ} + 18 0^{\circ} n

θ = 3 0^{\circ} + 18 0^{\circ} n, θ = 6 0^{\circ} + 18 0^{\circ} n

在

0 \leq θ < 18 0^{\circ}

范围内的解

θ = 3 0^{\circ}, θ = 6 0^{\circ}

作图

流行的例子

(sin(82))/(sin(x))=sqrt(5)

\frac{sin ( 8 2 ^{\circ} )}{sin ( x )} = 5

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

2 sin (x) = - 3

tan(θ)=(-12)/5 ,cot(θ)

tan (θ) = \frac{- 12}{5}, cot (θ)

sqrt(3)cos(1x)=-sin(1x)

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