What is the general solution for cos(θ/2)=sqrt((1+3/5)/2) ?

The general solution for cos(θ/2)=sqrt((1+3/5)/2) is θ=2*0.46364…+4pin,θ=4pi-2*0.46364…+4pin

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cos(θ/2)=sqrt((1+3/5)/2)

cos (\frac{θ}{2}) = \frac{1 + \frac{3}{5}}{2}

θ = 2 \cdot 0.46364 \dots + 4 πn, θ = 4 π - 2 \cdot 0.46364 \dots + 4 πn

Solution steps

cos (\frac{θ}{2}) = \frac{1 + \frac{3}{5}}{2}

Simplify

\frac{1 + \frac{3}{5}}{2} : \frac{2 5}{5}

Apply trig inverse properties

\frac{θ}{2} = arccos (\frac{2 5}{5}) + 2 πn, \frac{θ}{2} = 2 π - arccos (\frac{2 5}{5}) + 2 πn

Solve

\frac{θ}{2} = arccos (\frac{2 5}{5}) + 2 πn : θ = 2 arccos (\frac{2}{5}) + 4 πn

Solve

\frac{θ}{2} = 2 π - arccos (\frac{2 5}{5}) + 2 πn : θ = 4 π - 2 arccos (\frac{2}{5}) + 4 πn

θ = 2 arccos (\frac{2}{5}) + 4 πn, θ = 4 π - 2 arccos (\frac{2}{5}) + 4 πn

Show solutions in decimal form

θ = 2 \cdot 0.46364 \dots + 4 πn, θ = 4 π - 2 \cdot 0.46364 \dots + 4 πn

cos (θ) = \frac{35}{37}

- \frac{π}{2} = 1 - 2 sin (x)

sin (θ) = - 0.2761

sin (3 x) = sin (2 x + 50)

(r - c) = a 2 sin (θ) + d

What is the general solution for cos(θ/2)=sqrt((1+3/5)/2) ?
The general solution for cos(θ/2)=sqrt((1+3/5)/2) is θ=2*0.46364…+4pin,θ=4pi-2*0.46364…+4pin