What is the general solution for 1=lcos(x) ?

The general solution for 1=lcos(x) is x=arccos(1/l)+2pin,x=-arccos(1/l)+2pin

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1=lcos(x)

1 = l cos (x)

x = arccos (\frac{1}{l}) + 2 πn, x = - arccos (\frac{1}{l}) + 2 πn

Solution steps

1 = l cos (x)

Switch sides

l cos (x) = 1

Divide both sides by

l; l \neq = 0

l cos (x) = 1

Divide both sides by

l; l \neq = 0

\frac{l cos ( x )}{l} = \frac{1}{l}; l \neq = 0

Simplify

cos (x) = \frac{1}{l}; l \neq = 0

cos (x) = \frac{1}{l}; l \neq = 0

Apply trig inverse properties

cos (x) = \frac{1}{l}

General solutions for

cos (x) = \frac{1}{l}

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = - arccos (a) + 2 πn

x = arccos (\frac{1}{l}) + 2 πn, x = - arccos (\frac{1}{l}) + 2 πn

x = arccos (\frac{1}{l}) + 2 πn, x = - arccos (\frac{1}{l}) + 2 πn

tan (2 x) = cos (2 x), 0 \leq x \leq π

5 sin (2 x) = 5 cos (x), 0 \leq x \leq 2 π

x, f = cos (x) cos (h y)

tan (x) = - \frac{1}{10}

tan (4 x) \cdot cot (x + 60) = 1

What is the general solution for 1=lcos(x) ?
The general solution for 1=lcos(x) is x=arccos(1/l)+2pin,x=-arccos(1/l)+2pin