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受欢迎的三角函数 >

-2sqrt(2)=-4sin(2θ)

解答

- 22 = - 4 sin (2 θ)

解答

θ = \frac{π}{8} + πn, θ = \frac{3 π}{8} + πn

+1

度数

θ = 22. 5^{\circ} + 18 0^{\circ} n, θ = 67. 5^{\circ} + 18 0^{\circ} n

求解步骤

- 22 = - 4 sin (2 θ)

交换两边

- 4 sin (2 θ) = - 22

两边除以

- 4

- 4 sin (2 θ) = - 22

两边除以

- 4

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

化简

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

化简

\frac{- 4 sin ( 2 θ )}{- 4} : sin (2 θ)

\frac{- 4 sin ( 2 θ )}{- 4}

使用分式法则:

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

= \frac{4 sin ( 2 θ )}{4}

数字相除:

\frac{4}{4} = 1

= sin (2 θ)

化简

\frac{- 2 2}{- 4} : \frac{2}{2}

\frac{- 2 2}{- 4}

使用分式法则:

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

= \frac{2 2}{4}

约分:

2

= \frac{2}{2}

sin (2 θ) = \frac{2}{2}

sin (2 θ) = \frac{2}{2}

sin (2 θ) = \frac{2}{2}

sin (2 θ) = \frac{2}{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}

2 θ = \frac{π}{4} + 2 πn, 2 θ = \frac{3 π}{4} + 2 πn

2 θ = \frac{π}{4} + 2 πn, 2 θ = \frac{3 π}{4} + 2 πn

解

2 θ = \frac{π}{4} + 2 πn : θ = \frac{π}{8} + πn

2 θ = \frac{π}{4} + 2 πn

两边除以

2

2 θ = \frac{π}{4} + 2 πn

两边除以

2

\frac{2 θ}{2} = \frac{\frac{π}{4}}{2} + \frac{2 πn}{2}

化简

\frac{2 θ}{2} = \frac{\frac{π}{4}}{2} + \frac{2 πn}{2}

化简

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

数字相除:

\frac{2}{2} = 1

= θ

化简

\frac{\frac{π}{4}}{2} + \frac{2 πn}{2} : \frac{π}{8} + πn

\frac{\frac{π}{4}}{2} + \frac{2 πn}{2}

\frac{\frac{π}{4}}{2} = \frac{π}{8}

\frac{\frac{π}{4}}{2}

使用分式法则:

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

= \frac{π}{4 \cdot 2}

数字相乘:

4 \cdot 2 = 8

= \frac{π}{8}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

= \frac{π}{8} + πn

θ = \frac{π}{8} + πn

θ = \frac{π}{8} + πn

θ = \frac{π}{8} + πn

解

2 θ = \frac{3 π}{4} + 2 πn : θ = \frac{3 π}{8} + πn

2 θ = \frac{3 π}{4} + 2 πn

两边除以

2

2 θ = \frac{3 π}{4} + 2 πn

两边除以

2

\frac{2 θ}{2} = \frac{\frac{3 π}{4}}{2} + \frac{2 πn}{2}

化简

\frac{2 θ}{2} = \frac{\frac{3 π}{4}}{2} + \frac{2 πn}{2}

化简

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

数字相除:

\frac{2}{2} = 1

= θ

化简

\frac{\frac{3 π}{4}}{2} + \frac{2 πn}{2} : \frac{3 π}{8} + πn

\frac{\frac{3 π}{4}}{2} + \frac{2 πn}{2}

\frac{\frac{3 π}{4}}{2} = \frac{3 π}{8}

\frac{\frac{3 π}{4}}{2}

使用分式法则:

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

= \frac{3 π}{4 \cdot 2}

数字相乘:

4 \cdot 2 = 8

= \frac{3 π}{8}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

= \frac{3 π}{8} + πn

θ = \frac{3 π}{8} + πn

θ = \frac{3 π}{8} + πn

θ = \frac{3 π}{8} + πn

θ = \frac{π}{8} + πn, θ = \frac{3 π}{8} + πn

作图

流行的例子

8cos(x)=4cos^2(x)+3

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

(26)/(sin(88))=(21)/(sin(c))

\frac{26}{sin ( 8 8 ^{\circ} )} = \frac{21}{sin ( c )}

3 tan^{2} (3 x) = 1

6tan(x)-5csc(x)=0

6 tan (x) - 5 csc (x) = 0

3csc(x)-6=0,0<= x<= 360

3 csc (x) - 6 = 0, 0^{\circ} \leq x \leq 36 0^{\circ}