What is the general solution for -20cos(θ)+10=0 ?

The general solution for -20cos(θ)+10=0 is θ= pi/3+2pin,θ=(5pi)/3+2pin

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-20cos(θ)+10=0

- 20 cos (θ) + 10 = 0

θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

Solution steps

- 20 cos (θ) + 10 = 0

Move

10

to the right side

- 20 cos (θ) = - 10

Divide both sides by

- 20

cos (θ) = \frac{1}{2}

General solutions for

cos (θ) = \frac{1}{2}

θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

sin (x) - 1 = - 1

arctan (5 w) = 1

14 cos (x^{2}) sin (e^{x}) = 0

- 20 sin (4 x) = 0

x, y = 2 sin (\frac{π}{4} x)

What is the general solution for -20cos(θ)+10=0 ?
The general solution for -20cos(θ)+10=0 is θ= pi/3+2pin,θ=(5pi)/3+2pin