What is the general solution for cos^2(θ)+(20^2cos(θ))/(9.81(1))-1=0 ?

The general solution for cos^2(θ)+(20^2cos(θ))/(9.81(1))-1=0 is θ=1.54628…+2pin,θ=2pi-1.54628…+2pin

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cos^2(θ)+(20^2cos(θ))/(9.81(1))-1=0

cos^{2} (θ) + \frac{2 0 ^{2} cos ( θ )}{9.81 ( 1 )} - 1 = 0

θ = 1.54628 \dots + 2 πn, θ = 2 π - 1.54628 \dots + 2 πn

Solution steps

cos^{2} (θ) + \frac{2 0 ^{2} cos ( θ )}{9.81 ( 1 )} - 1 = 0

Solve by substitution

cos (θ) = \frac{- 400 + 160384.9444}{19.62}, cos (θ) = \frac{- 400 - 160384.9444}{19.62}

cos (θ) = \frac{- 400 + 160384.9444}{19.62} : θ = arccos (\frac{- 400 + 160384.9444}{19.62}) + 2 πn, θ = 2 π - arccos (\frac{- 400 + 160384.9444}{19.62}) + 2 πn

cos (θ) = \frac{- 400 - 160384.9444}{19.62} :

No Solution

Combine all the solutions

θ = arccos (\frac{- 400 + 160384.9444}{19.62}) + 2 πn, θ = 2 π - arccos (\frac{- 400 + 160384.9444}{19.62}) + 2 πn

Show solutions in decimal form

θ = 1.54628 \dots + 2 πn, θ = 2 π - 1.54628 \dots + 2 πn

3 sin (θ) + 5 = 6

\frac{9}{73} + sin^{2} (θ) = 1

x, - 2 sin (2 x) - 2 sin (4 x) = 0

6.15 = 220 sin (377 t)

x, 3 sec (5 x) - 2 = 0

What is the general solution for cos^2(θ)+(20^2cos(θ))/(9.81(1))-1=0 ?
The general solution for cos^2(θ)+(20^2cos(θ))/(9.81(1))-1=0 is θ=1.54628…+2pin,θ=2pi-1.54628…+2pin