What is the general solution for arctan(x)=((2*1.8))/([(3170+1.7)-1]) ?

The general solution for arctan(x)=((2*1.8))/([(3170+1.7)-1]) is x=tan(4/3523)

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arctan(x)=((2*1.8))/([(3170+1.7)-1])

arctan (x) = \frac{( 2 \cdot 1.8 )}{[ ( 3170 + 1.7 ) - 1 ]}

x = tan (\frac{4}{3523})

Solution steps

arctan (x) = \frac{( 2 \cdot 1.8 )}{[ ( 3170 + 1.7 ) - 1 ]}

Simplify

\frac{2 \cdot 1.8}{3170 + 1.7 - 1} : \frac{3.6}{3170.7}

\frac{2 \cdot 1.8}{[ 3170 + 1.7 - 1 ]}

Multiply the numbers:

2 \cdot 1.8 = 3.6

= \frac{3.6}{3170 + 1.7 - 1}

Add/Subtract the numbers:

3170 + 1.7 - 1 = 3170.7

= \frac{3.6}{3170.7}

Apply trig inverse properties

arctan (x) = \frac{3.6}{3170.7}

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

x = tan (\frac{3.6}{3170.7})

tan (\frac{3.6}{3170.7}) = tan (\frac{\frac{18}{5}}{\frac{31707}{10}})

tan (\frac{3.6}{3170.7})

x = tan (\frac{\frac{18}{5}}{\frac{31707}{10}})

x = tan (\frac{\frac{18}{5}}{\frac{31707}{10}})

tan (\frac{\frac{18}{5}}{\frac{31707}{10}}) = tan (\frac{4}{3523})

tan (\frac{\frac{18}{5}}{\frac{31707}{10}})

= tan (\frac{4}{3523})

x = tan (\frac{4}{3523})

cos (4 x) + sin (\frac{x}{2}) = 0

2 sin^{2} (θ) - 5 sin (θ) = 3

cos (3 x - \frac{π}{4}) = \frac{1}{2}, 0 \leq x \leq π

4 sin^{2} (x) = 3, 0 \leq x \leq 10 8^{\circ}

3 sin (x) + 3 cos (x) = 0

What is the general solution for arctan(x)=((2*1.8))/([(3170+1.7)-1]) ?
The general solution for arctan(x)=((2*1.8))/([(3170+1.7)-1]) is x=tan(4/3523)