What is the general solution for solvefor θ,w=fdcos(θ) ?

The general solution for solvefor θ,w=fdcos(θ) is θ=arccos(w/(fd))+2pin,θ=-arccos(w/(fd))+2pin

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solvefor θ,w=fdcos(θ)

solve for

θ, w = fd cos (θ)

θ = arccos (\frac{w}{fd}) + 2 πn, θ = - arccos (\frac{w}{fd}) + 2 πn

Solution steps

w = fd cos (θ)

Switch sides

fd cos (θ) = w

Divide both sides by

fd; fd \neq = 0

fd cos (θ) = w

Divide both sides by

fd; fd \neq = 0

\frac{fd cos ( θ )}{fd} = \frac{w}{fd}; fd \neq = 0

Simplify

cos (θ) = \frac{w}{fd}; fd \neq = 0

cos (θ) = \frac{w}{fd}; fd \neq = 0

Apply trig inverse properties

cos (θ) = \frac{w}{fd}

General solutions for

cos (θ) = \frac{w}{fd}

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

θ = arccos (\frac{w}{fd}) + 2 πn, θ = - arccos (\frac{w}{fd}) + 2 πn

θ = arccos (\frac{w}{fd}) + 2 πn, θ = - arccos (\frac{w}{fd}) + 2 πn

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7 cos (2 x) = 7 sin^{2} (x) + 5

What is the general solution for solvefor θ,w=fdcos(θ) ?
The general solution for solvefor θ,w=fdcos(θ) is θ=arccos(w/(fd))+2pin,θ=-arccos(w/(fd))+2pin