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受欢迎的三角函数 >

n=(2tan^3(2θ)-1)^{1/2}

解答

n = (2 tan^{3} (2 θ) - 1)^{\frac{1}{2}}

解答

θ = \frac{arctan ( 3 \frac{n ^{2} + 1}{2} )}{2} + \frac{πk}{2}

求解步骤

n = (2 tan^{3} (2 θ) - 1)^{\frac{1}{2}}

交换两边

(2 tan^{3} (2 θ) - 1)^{\frac{1}{2}} = n

两边进行平方:

2 tan^{3} (2 θ) - 1 = n^{2}

(2 tan^{3} (2 θ) - 1)^{\frac{1}{2}} = n

((2 tan^{3} (2 θ) - 1)^{\frac{1}{2}})^{2} = n^{2}

展开

((2 tan^{3} (2 θ) - 1)^{\frac{1}{2}})^{2} : 2 tan^{3} (2 θ) - 1

((2 tan^{3} (2 θ) - 1)^{\frac{1}{2}})^{2}

使用指数法则:

(a^{b})^{c} = a^{b c}

= (2 tan^{3} (2 θ) - 1)^{\frac{1}{2} \cdot 2}

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

\frac{1}{2} \cdot 2

分式相乘:

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

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

约分:

2

= 1

= 2 tan^{3} (2 θ) - 1

2 tan^{3} (2 θ) - 1 = n^{2}

2 tan^{3} (2 θ) - 1 = n^{2}

解

2 tan^{3} (2 θ) - 1 = n^{2} : tan (2 θ) = 3 \frac{n ^{2} + 1}{2}

2 tan^{3} (2 θ) - 1 = n^{2}

将

1

到右边

2 tan^{3} (2 θ) - 1 = n^{2}

两边加上

1

2 tan^{3} (2 θ) - 1 + 1 = n^{2} + 1

化简

2 tan^{3} (2 θ) = n^{2} + 1

2 tan^{3} (2 θ) = n^{2} + 1

两边除以

2

2 tan^{3} (2 θ) = n^{2} + 1

两边除以

2

\frac{2 tan ^{3} ( 2 θ )}{2} = \frac{n ^{2}}{2} + \frac{1}{2}

化简

\frac{2 tan ^{3} ( 2 θ )}{2} = \frac{n ^{2}}{2} + \frac{1}{2}

化简

\frac{2 tan ^{3} ( 2 θ )}{2} : tan^{3} (2 θ)

\frac{2 tan ^{3} ( 2 θ )}{2}

数字相除:

\frac{2}{2} = 1

= tan^{3} (2 θ)

化简

\frac{n ^{2}}{2} + \frac{1}{2} : \frac{n ^{2} + 1}{2}

\frac{n ^{2}}{2} + \frac{1}{2}

使用法则

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

= \frac{n ^{2} + 1}{2}

tan^{3} (2 θ) = \frac{n ^{2} + 1}{2}

tan^{3} (2 θ) = \frac{n ^{2} + 1}{2}

tan^{3} (2 θ) = \frac{n ^{2} + 1}{2}

对于

x^{n} = f (a)

，n 为奇数，解为

x = n f (a)

tan (2 θ) = 3 \frac{n ^{2} + 1}{2}

tan (2 θ) = 3 \frac{n ^{2} + 1}{2}

验证解:

tan (2 θ) = 3 \frac{n ^{2} + 1}{2} {n \geq 0}

将它们代入

(2 tan^{3} (2 θ) - 1)^{\frac{1}{2}} = n

检验解是否符合

去除与方程不符的解。

代入

tan (2 θ) = 3 \frac{n ^{2} + 1}{2} : 2 (3 \frac{n ^{2} + 1}{2})^{3} - 1^{\frac{1}{2}} = n \Rightarrow n \geq 0

2 (3 \frac{n ^{2} + 1}{2})^{3} - 1^{\frac{1}{2}} = n

两边进行平方:

n^{2} = n^{2}

2 (3 \frac{n ^{2} + 1}{2})^{3} - 1^{\frac{1}{2}} = n

2 (3 \frac{n ^{2} + 1}{2})^{3} - 1^{\frac{1}{2}}^{2} = n^{2}

展开

2 (3 \frac{n ^{2} + 1}{2})^{3} - 1^{\frac{1}{2}}^{2} : n^{2}

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

使用指数法则:

(a^{b})^{c} = a^{b c}

= 2 (3 \frac{n ^{2} + 1}{2})^{3} - 1^{\frac{1}{2} \cdot 2}

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

\frac{1}{2} \cdot 2

分式相乘:

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

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

约分:

2

= 1

= 2 (3 \frac{n ^{2} + 1}{2})^{3} - 1

展开

2 (3 \frac{n ^{2} + 1}{2})^{3} - 1 : n^{2}

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

2 (3 \frac{n ^{2} + 1}{2})^{3} = n^{2} + 1

2 (3 \frac{n ^{2} + 1}{2})^{3}

(3 \frac{n ^{2} + 1}{2})^{3} = \frac{n ^{2} + 1}{2}

(3 \frac{n ^{2} + 1}{2})^{3}

使用指数法则:

(a^{b})^{c} = a^{b c}

= (\frac{n ^{2} + 1}{2})^{\frac{1}{3} \cdot 3}

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

\frac{1}{3} \cdot 3

分式相乘:

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

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

约分:

3

= 1

= \frac{n ^{2} + 1}{2}

= 2 \cdot \frac{n ^{2} + 1}{2}

分式相乘:

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

= \frac{( n ^{2} + 1 ) \cdot 2}{2}

约分:

2

= n^{2} + 1

= n^{2} + 1 - 1

1 - 1 = 0

= n^{2}

= n^{2}

n^{2} = n^{2}

n^{2} = n^{2}

两侧相等

对所有 n 为真

验证解:

n < 0

假

, n = 0

真

, n > 0

真

2 (3 \frac{n ^{2} + 1}{2})^{3} - 1^{\frac{1}{2}} = n

将定义域区间与解区间合并:

对所有 n 为真

找到函数区间:

n < 0, n = 0, n > 0

2 (3 \frac{n ^{2} + 1}{2})^{3} - 1^{\frac{1}{2}} = n

找到偶数根参量零值:

解

2 3 \frac{n ^{2} + 1}{2}^{3} - 1 = 0 : n = 0

2 (3 \frac{n ^{2} + 1}{2})^{3} - 1 = 0

因式分解

2 (3 \frac{n ^{2} + 1}{2})^{3} - 1 : (2^{\frac{1}{3}} 3 \frac{n ^{2} + 1}{2} - 1) (2^{\frac{2}{3}} (\frac{n ^{2} + 1}{2})^{\frac{2}{3}} + 2^{\frac{1}{3}} 3 \frac{n ^{2} + 1}{2} + 1)

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

将

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

改写为

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

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

将

2

改写为

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

= (2^{\frac{1}{3}})^{3} (3 \frac{n ^{2} + 1}{2})^{3} - 1

将

1

改写为

1^{3}

= (2^{\frac{1}{3}})^{3} (3 \frac{n ^{2} + 1}{2})^{3} - 1^{3}

使用指数法则:

a^{m} b^{m} = (ab)^{m}

(2^{\frac{1}{3}})^{3} (3 \frac{n ^{2} + 1}{2})^{3} = (2^{\frac{1}{3}} 3 \frac{n ^{2} + 1}{2})^{3}

= (2^{\frac{1}{3}} 3 \frac{n ^{2} + 1}{2})^{3} - 1^{3}

= (2^{\frac{1}{3}} 3 \frac{n ^{2} + 1}{2})^{3} - 1^{3}

使用立方差公式:

x^{3} - y^{3} = (x - y) (x^{2} + x y + y^{2})

(2^{\frac{1}{3}} 3 \frac{n ^{2} + 1}{2})^{3} - 1^{3} = (2^{\frac{1}{3}} 3 \frac{n ^{2} + 1}{2} - 1) (2^{\frac{1}{3}})^{2} (3 \frac{n ^{2} + 1}{2})^{2} + 2^{\frac{1}{3}} 3 \frac{n ^{2} + 1}{2} + 1

= (2^{\frac{1}{3}} 3 \frac{n ^{2} + 1}{2} - 1) (2^{\frac{1}{3}})^{2} (3 \frac{n ^{2} + 1}{2})^{2} + 2^{\frac{1}{3}} 3 \frac{n ^{2} + 1}{2} + 1

整理后得

= (2^{\frac{1}{3}} 3 \frac{n ^{2} + 1}{2} - 1) 2^{\frac{2}{3}} (3 \frac{n ^{2} + 1}{2})^{2} + 2^{\frac{1}{3}} 3 \frac{n ^{2} + 1}{2} + 1

(3 \frac{n ^{2} + 1}{2})^{2} = \frac{n ^{2} + 1}{2}^{\frac{2}{3}}

(3 \frac{n ^{2} + 1}{2})^{2}

使用指数法则:

(a^{b})^{c} = a^{b c}

= (\frac{n ^{2} + 1}{2})^{\frac{1}{3} \cdot 2}

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

\frac{1}{3} \cdot 2

分式相乘:

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

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

数字相乘:

1 \cdot 2 = 2

= \frac{2}{3}

= (\frac{n ^{2} + 1}{2})^{\frac{2}{3}}

= (2^{\frac{1}{3}} 3 \frac{n ^{2} + 1}{2} - 1) (2^{\frac{2}{3}} (\frac{n ^{2} + 1}{2})^{\frac{2}{3}} + 2^{\frac{1}{3}} 3 \frac{n ^{2} + 1}{2} + 1)

(2^{\frac{1}{3}} 3 \frac{n ^{2} + 1}{2} - 1) (2^{\frac{2}{3}} (\frac{n ^{2} + 1}{2})^{\frac{2}{3}} + 2^{\frac{1}{3}} 3 \frac{n ^{2} + 1}{2} + 1) = 0

使用零因数法则： If

ab = 0

then

a = 0

or

b = 0

2^{\frac{1}{3}} 3 \frac{n ^{2} + 1}{2} - 1 = 0 or 2^{\frac{2}{3}} (\frac{n ^{2} + 1}{2})^{\frac{2}{3}} + 2^{\frac{1}{3}} 3 \frac{n ^{2} + 1}{2} + 1 = 0

解

2^{\frac{1}{3}} 3 \frac{n ^{2} + 1}{2} - 1 = 0 : n = 0

2^{\frac{1}{3}} 3 \frac{n ^{2} + 1}{2} - 1 = 0

将

1

到右边

2^{\frac{1}{3}} 3 \frac{n ^{2} + 1}{2} - 1 = 0

两边加上

1

2^{\frac{1}{3}} 3 \frac{n ^{2} + 1}{2} - 1 + 1 = 0 + 1

化简

2^{\frac{1}{3}} 3 \frac{n ^{2} + 1}{2} = 1

2^{\frac{1}{3}} 3 \frac{n ^{2} + 1}{2} = 1

两边除以

2^{\frac{1}{3}}

2^{\frac{1}{3}} 3 \frac{n ^{2} + 1}{2} = 1

两边除以

2^{\frac{1}{3}}

\frac{2 ^{\frac{1}{3}} 3 \frac{n ^{2} + 1}{2}}{2 ^{\frac{1}{3}}} = \frac{1}{2 ^{\frac{1}{3}}}

化简

3 \frac{n ^{2} + 1}{2} = \frac{1}{2 ^{\frac{1}{3}}}

3 \frac{n ^{2} + 1}{2} = \frac{1}{2 ^{\frac{1}{3}}}

对方程式两边

3

次方:

\frac{n ^{2} + 1}{2} = \frac{1}{2}

3 \frac{n ^{2} + 1}{2} = \frac{1}{2 ^{\frac{1}{3}}}

(3 \frac{n ^{2} + 1}{2})^{3} = (\frac{1}{2 ^{\frac{1}{3}}})^{3}

展开

(3 \frac{n ^{2} + 1}{2})^{3} : \frac{n ^{2} + 1}{2}

(3 \frac{n ^{2} + 1}{2})^{3}

使用指数法则:

(a^{b})^{c} = a^{b c}

= (\frac{n ^{2} + 1}{2})^{\frac{1}{3} \cdot 3}

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

\frac{1}{3} \cdot 3

分式相乘:

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

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

约分:

3

= 1

= \frac{n ^{2} + 1}{2}

展开

(\frac{1}{2 ^{\frac{1}{3}}})^{3} : \frac{1}{2}

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

使用指数法则:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{1 ^{3}}{( 2 ^{\frac{1}{3}} ) ^{3}}

(2^{\frac{1}{3}})^{3} : 2

使用指数法则:

(a^{b})^{c} = a^{b c}

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

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

\frac{1}{3} \cdot 3

分式相乘:

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

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

约分:

3

= 1

= 2

= \frac{1 ^{3}}{2}

使用法则

1^{a} = 1

1^{3} = 1

= \frac{1}{2}

\frac{n ^{2} + 1}{2} = \frac{1}{2}

\frac{n ^{2} + 1}{2} = \frac{1}{2}

解

\frac{n ^{2} + 1}{2} = \frac{1}{2} : n = 0

\frac{n ^{2} + 1}{2} = \frac{1}{2}

在两边乘以

2

\frac{n ^{2} + 1}{2} = \frac{1}{2}

在两边乘以

2

\frac{2 ( n ^{2} + 1 )}{2} = \frac{1 \cdot 2}{2}

化简

\frac{2 ( n ^{2} + 1 )}{2} = \frac{1 \cdot 2}{2}

化简

\frac{2 ( n ^{2} + 1 )}{2} : n^{2} + 1

\frac{2 ( n ^{2} + 1 )}{2}

数字相除:

\frac{2}{2} = 1

= n^{2} + 1

化简

\frac{1 \cdot 2}{2} : 1

\frac{1 \cdot 2}{2}

数字相乘:

1 \cdot 2 = 2

= \frac{2}{2}

使用法则

\frac{a}{a} = 1

= 1

n^{2} + 1 = 1

n^{2} + 1 = 1

n^{2} + 1 = 1

将

1

到右边

n^{2} + 1 = 1

两边减去

1

n^{2} + 1 - 1 = 1 - 1

化简

n^{2} = 0

n^{2} = 0

使用法则

x^{n} = 0 \Rightarrow x = 0

n = 0

n = 0

验证解:

n = 0

真

将它们代入

2^{\frac{1}{3}} 3 \frac{n ^{2} + 1}{2} - 1 = 0

检验解是否符合

去除与方程不符的解。

代入

n = 0 :

真

2^{\frac{1}{3}} 3 \frac{0 ^{2} + 1}{2} - 1 = 0

2^{\frac{1}{3}} 3 \frac{0 ^{2} + 1}{2} - 1 = 0

2^{\frac{1}{3}} 3 \frac{0 ^{2} + 1}{2} - 1

使用法则

0^{a} = 0

0^{2} = 0

= 2^{\frac{1}{3}} 3 \frac{0 + 1}{2} - 1

2^{\frac{1}{3}} 3 \frac{0 + 1}{2} = 1

2^{\frac{1}{3}} 3 \frac{0 + 1}{2}

数字相加:

0 + 1 = 1

= 2^{\frac{1}{3}} 3 \frac{1}{2}

使用指数法则:

a^{m} b^{m} = (ab)^{m}

2^{\frac{1}{3}} 3 \frac{1}{2} = (2 \cdot \frac{1}{2})^{\frac{1}{3}}

= (2 \cdot \frac{1}{2})^{\frac{1}{3}}

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

2 \cdot \frac{1}{2}

分式相乘:

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

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

约分:

2

= 1

= 1^{\frac{1}{3}}

使用法则

1^{a} = 1

= 1

= 1 - 1

数字相减:

1 - 1 = 0

= 0

0 = 0

真

解是

n = 0

解

2^{\frac{2}{3}} (\frac{n ^{2} + 1}{2})^{\frac{2}{3}} + 2^{\frac{1}{3}} 3 \frac{n ^{2} + 1}{2} + 1 = 0 : n \in R

无解

2^{\frac{2}{3}} (\frac{n ^{2} + 1}{2})^{\frac{2}{3}} + 2^{\frac{1}{3}} 3 \frac{n ^{2} + 1}{2} + 1 = 0

使用以下的指数特性

: a^{\frac{n}{m}} = (m a)^{n}

(\frac{n ^{2} + 1}{2})^{\frac{2}{3}} = (3 \frac{n ^{2} + 1}{2})^{2}

2^{\frac{2}{3}} (3 \frac{n ^{2} + 1}{2})^{2} + 2^{\frac{1}{3}} 3 \frac{n ^{2} + 1}{2} + 1 = 0

用

3 \frac{n ^{2} + 1}{2} = u

改写方程式

2^{\frac{2}{3}} u^{2} + 2^{\frac{1}{3}} u + 1 = 0

解

2^{\frac{2}{3}} u^{2} + 2^{\frac{1}{3}} u + 1 = 0 : u \in R

无解

2^{\frac{2}{3}} u^{2} + 2^{\frac{1}{3}} u + 1 = 0

判别式

2^{\frac{2}{3}} u^{2} + 2^{\frac{1}{3}} u + 1 = 0 : - 3 \cdot 2^{\frac{2}{3}}

2^{\frac{2}{3}} u^{2} + 2^{\frac{1}{3}} u + 1 = 0

对于

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

形式的二次方程，根判别式为

b^{2} - 4 a c

若

a = 2^{\frac{2}{3}}, b = 2^{\frac{1}{3}}, c = 1 : (2^{\frac{1}{3}})^{2} - 4 \cdot 2^{\frac{2}{3}} \cdot 1

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

展开

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

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

(2^{\frac{1}{3}})^{2} = 2^{\frac{2}{3}}

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

使用指数法则:

(a^{b})^{c} = a^{b c}

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

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

\frac{1}{3} \cdot 2

分式相乘:

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

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

数字相乘:

1 \cdot 2 = 2

= \frac{2}{3}

= 2^{\frac{2}{3}}

4 \cdot 2^{\frac{2}{3}} \cdot 1 = 4 \cdot 2^{\frac{2}{3}}

4 \cdot 2^{\frac{2}{3}} \cdot 1

数字相乘:

4 \cdot 1 = 4

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

= 2^{\frac{2}{3}} - 4 \cdot 2^{\frac{2}{3}}

同类项相加:

2^{\frac{2}{3}} - 4 \cdot 2^{\frac{2}{3}} = - 3 \cdot 2^{\frac{2}{3}}

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

- 3 \cdot 2^{\frac{2}{3}}

判别式在

u

内不能为负

\in R

解是

u \in R 无解

n \in R 无解

n = 0

验证解:

n = 0

真

将它们代入

2 3 \frac{n ^{2} + 1}{2}^{3} - 1 = 0

检验解是否符合

去除与方程不符的解。

代入

n = 0 :

真

2 (3 \frac{0 ^{2} + 1}{2})^{3} - 1 = 0

2 (3 \frac{0 ^{2} + 1}{2})^{3} - 1 = 0

2 (3 \frac{0 ^{2} + 1}{2})^{3} - 1

使用法则

0^{a} = 0

0^{2} = 0

= 2 (3 \frac{0 + 1}{2})^{3} - 1

2 (3 \frac{0 + 1}{2})^{3} = 1

2 (3 \frac{0 + 1}{2})^{3}

(3 \frac{0 + 1}{2})^{3} = \frac{1}{2}

(3 \frac{0 + 1}{2})^{3}

使用指数法则:

(a^{b})^{c} = a^{b c}

= (\frac{0 + 1}{2})^{\frac{1}{3} \cdot 3}

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

\frac{1}{3} \cdot 3

分式相乘:

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

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

约分:

3

= 1

= \frac{0 + 1}{2}

数字相加:

0 + 1 = 1

= \frac{1}{2}

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

分式相乘:

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

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

约分:

2

= 1

= 1 - 1

数字相减:

1 - 1 = 0

= 0

0 = 0

真

解是

n = 0

n = 0

区间围绕零点来定义:

n < 0, n = 0, n > 0

将区间与定义域合并

n < 0, n = 0, n > 0

将它们代入

(2 3 \frac{n ^{2} + 1}{2}^{3} - 1)^{\frac{1}{2}} = n

检验解是否符合

去除与方程不符的解。

代入

n < 0 : (2 3 \frac{n ^{2} + 1}{2}^{3} - 1)^{\frac{1}{2}} \neq = n \Rightarrow

假

解是

n \geq 0

解是

tan (2 θ) = 3 \frac{n ^{2} + 1}{2} {n \geq 0}

使用反三角函数性质

tan (2 θ) = 3 \frac{n ^{2} + 1}{2}

tan (2 θ) = 3 \frac{n ^{2} + 1}{2}

的通解

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

2 θ = arctan (3 \frac{n ^{2} + 1}{2}) + πk

2 θ = arctan (3 \frac{n ^{2} + 1}{2}) + πk

解

2 θ = arctan (3 \frac{n ^{2} + 1}{2}) + πk : θ = \frac{arctan ( 3 \frac{n ^{2} + 1}{2} )}{2} + \frac{πk}{2}

2 θ = arctan (3 \frac{n ^{2} + 1}{2}) + πk

两边除以

2

2 θ = arctan (3 \frac{n ^{2} + 1}{2}) + πk

两边除以

2

\frac{2 θ}{2} = \frac{arctan ( 3 \frac{n ^{2} + 1}{2} )}{2} + \frac{πk}{2}

化简

θ = \frac{arctan ( 3 \frac{n ^{2} + 1}{2} )}{2} + \frac{πk}{2}

θ = \frac{arctan ( 3 \frac{n ^{2} + 1}{2} )}{2} + \frac{πk}{2}

θ = \frac{arctan ( 3 \frac{n ^{2} + 1}{2} )}{2} + \frac{πk}{2}

作图

流行的例子

0 = 2 sin (x + 1) + 1

tan (θ) \cdot 10 = 10

4cos(x)+tan(45)=0.6

4 cos (x) + tan (4 5^{\circ}) = 0.6

sin(5x+8)=cos(9x-16)

sin (5 x + 8) = cos (9 x - 16)

cos(x)=(sqrt(125))/(sqrt(174))

cos (x^{\circ}) = \frac{125}{174}