What is the general solution for ω=360lb*5sec(x) ?

The general solution for ω=360lb*5sec(x) is x=arcsec(ω/(1800lb))+2pin,x=-arcsec(ω/(1800lb))+2pin

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ω=360lb*5sec(x)

ω = 360 l b \cdot 5 sec (x)

x = arcsec (\frac{ω}{1800 l b}) + 2 πn, x = - arcsec (\frac{ω}{1800 l b}) + 2 πn

Solution steps

ω = 360 l b \cdot 5 sec (x)

Switch sides

360 l b \cdot 5 sec (x) = ω

Divide both sides by

1800 l b; 1800 l b \neq = 0

360 l b \cdot 5 sec (x) = ω

Divide both sides by

1800 l b; 1800 l b \neq = 0

\frac{360 l b \cdot 5 sec ( x )}{1800 l b} = \frac{ω}{1800 l b}; 1800 l b \neq = 0

Simplify

sec (x) = \frac{ω}{1800 l b}; 1800 l b \neq = 0

sec (x) = \frac{ω}{1800 l b}; 1800 l b \neq = 0

Apply trig inverse properties

sec (x) = \frac{ω}{1800 l b}

General solutions for

sec (x) = \frac{ω}{1800 l b}

sec (x) = a \Rightarrow x = arcsec (a) + 2 πn, x = - arcsec (a) + 2 πn

x = arcsec (\frac{ω}{1800 l b}) + 2 πn, x = - arcsec (\frac{ω}{1800 l b}) + 2 πn

x = arcsec (\frac{ω}{1800 l b}) + 2 πn, x = - arcsec (\frac{ω}{1800 l b}) + 2 πn

u, x = 2 cos (u) cos (v)

cos (θ) = cos (3 θ)

sec (x + \frac{π}{4}) = - 2

cot (3 θ) = 3

tan (x) = \frac{45}{35}

What is the general solution for ω=360lb*5sec(x) ?
The general solution for ω=360lb*5sec(x) is x=arcsec(ω/(1800lb))+2pin,x=-arcsec(ω/(1800lb))+2pin