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受欢迎的三角函数 >

8cos(2t)-14sin(2t)=0

解答

8 cos (2 t) - 14 sin (2 t) = 0

解答

t = \frac{0.51914 \dots}{2} + \frac{πn}{2}

+1

度数

t = 14.87244 \dots^{\circ} + 9 0^{\circ} n

求解步骤

8 cos (2 t) - 14 sin (2 t) = 0

使用三角恒等式改写

8 cos (2 t) - 14 sin (2 t) = 0

在两边除以

cos (2 t), cos (2 t) \neq = 0

\frac{8 cos ( 2 t ) - 14 sin ( 2 t )}{cos ( 2 t )} = \frac{0}{cos ( 2 t )}

化简

8 - \frac{14 sin ( 2 t )}{cos ( 2 t )} = 0

使用基本三角恒等式:

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

8 - 14 tan (2 t) = 0

8 - 14 tan (2 t) = 0

将

8

到右边

8 - 14 tan (2 t) = 0

两边减去

8

8 - 14 tan (2 t) - 8 = 0 - 8

化简

- 14 tan (2 t) = - 8

- 14 tan (2 t) = - 8

两边除以

- 14

- 14 tan (2 t) = - 8

两边除以

- 14

\frac{- 14 tan ( 2 t )}{- 14} = \frac{- 8}{- 14}

化简

\frac{- 14 tan ( 2 t )}{- 14} = \frac{- 8}{- 14}

化简

\frac{- 14 tan ( 2 t )}{- 14} : tan (2 t)

\frac{- 14 tan ( 2 t )}{- 14}

使用分式法则:

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

= \frac{14 tan ( 2 t )}{14}

数字相除:

\frac{14}{14} = 1

= tan (2 t)

化简

\frac{- 8}{- 14} : \frac{4}{7}

\frac{- 8}{- 14}

使用分式法则:

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

= \frac{8}{14}

约分:

2

= \frac{4}{7}

tan (2 t) = \frac{4}{7}

tan (2 t) = \frac{4}{7}

tan (2 t) = \frac{4}{7}

使用反三角函数性质

tan (2 t) = \frac{4}{7}

tan (2 t) = \frac{4}{7}

的通解

tan (x) = a \Rightarrow x = arctan (a) + πn

2 t = arctan (\frac{4}{7}) + πn

2 t = arctan (\frac{4}{7}) + πn

解

2 t = arctan (\frac{4}{7}) + πn : t = \frac{arctan ( \frac{4}{7} )}{2} + \frac{πn}{2}

2 t = arctan (\frac{4}{7}) + πn

两边除以

2

2 t = arctan (\frac{4}{7}) + πn

两边除以

2

\frac{2 t}{2} = \frac{arctan ( \frac{4}{7} )}{2} + \frac{πn}{2}

化简

t = \frac{arctan ( \frac{4}{7} )}{2} + \frac{πn}{2}

t = \frac{arctan ( \frac{4}{7} )}{2} + \frac{πn}{2}

t = \frac{arctan ( \frac{4}{7} )}{2} + \frac{πn}{2}

以小数形式表示解

t = \frac{0.51914 \dots}{2} + \frac{πn}{2}

作图

流行的例子

1-sqrt(2)sin(θ)=0

1 - 2 sin (θ) = 0

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

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

2cos^2(θ)+4sin(θ)-13=9sin(θ)-9

2 cos^{2} (θ) + 4 sin (θ) - 13 = 9 sin (θ) - 9

cot (θ) = - \frac{5}{4}

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