What is the general solution for cos(θ)a= pi/4 ?

The general solution for cos(θ)a= pi/4 is θ=arccos(pi/(4a))+2pin,θ=-arccos(pi/(4a))+2pin

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cos(θ)a= pi/4

cos (θ) a = \frac{π}{4}

θ = arccos (\frac{π}{4 a}) + 2 πn, θ = - arccos (\frac{π}{4 a}) + 2 πn

Solution steps

cos (θ) a = \frac{π}{4}

Divide both sides by

a; a \neq = 0

cos (θ) a = \frac{π}{4}

Divide both sides by

a; a \neq = 0

\frac{cos ( θ ) a}{a} = \frac{\frac{π}{4}}{a}; a \neq = 0

Simplify

\frac{cos ( θ ) a}{a} = \frac{\frac{π}{4}}{a}

Simplify

\frac{cos ( θ ) a}{a} : cos (θ)

\frac{cos ( θ ) a}{a}

Cancel the common factor:

a

= cos (θ)

Simplify

\frac{\frac{π}{4}}{a} : \frac{π}{4 a}

\frac{\frac{π}{4}}{a}

Apply the fraction rule:

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

= \frac{π}{4 a}

cos (θ) = \frac{π}{4 a}; a \neq = 0

cos (θ) = \frac{π}{4 a}; a \neq = 0

cos (θ) = \frac{π}{4 a}; a \neq = 0

Apply trig inverse properties

cos (θ) = \frac{π}{4 a}

General solutions for

cos (θ) = \frac{π}{4 a}

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

θ = arccos (\frac{π}{4 a}) + 2 πn, θ = - arccos (\frac{π}{4 a}) + 2 πn

θ = arccos (\frac{π}{4 a}) + 2 πn, θ = - arccos (\frac{π}{4 a}) + 2 πn

cos (x) = \frac{59}{95}

tan^{2} (θ) = \frac{3}{8}

sin (a) = \frac{5}{9}

sin (x - 2 0^{\circ}) = \frac{1}{2}

tan^{2} (x) + 2 sec^{2} (x) = 3

What is the general solution for cos(θ)a= pi/4 ?
The general solution for cos(θ)a= pi/4 is θ=arccos(pi/(4a))+2pin,θ=-arccos(pi/(4a))+2pin