What is the general solution for cos(θ)=cos(1) ?

The general solution for cos(θ)=cos(1) is θ=1+2pin,θ=2pi-1+2pin

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cos(θ)=cos(1)

cos (θ) = cos (1)

θ = 1 + 2 πn, θ = 2 π - 1 + 2 πn

Solution steps

cos (θ) = cos (1)

Apply trig inverse properties

θ = arccos (cos (1)) + 2 πn, θ = 2 π - arccos (cos (1)) + 2 πn

Show solutions in decimal form

θ = 1 + 2 πn, θ = 2 π - 1 + 2 πn

tan (\frac{x}{2}) = \pm \frac{1 - cos ( x )}{1 + cos ( x )}

sin^{2} (x) - cos^{2} (x) = 0.5

sec (x) - tan (x) = 1

7 cos^{2} (θ) + 6 cos (θ) - 12 = 7 cos (θ) - 6

sin^{2} (x) = sin (x) + 1 - sin^{2} (x)

What is the general solution for cos(θ)=cos(1) ?
The general solution for cos(θ)=cos(1) is θ=1+2pin,θ=2pi-1+2pin