What is the general solution for 3cos(θ)+1=1 ?

The general solution for 3cos(θ)+1=1 is θ= pi/2+2pin,θ=(3pi)/2+2pin

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3cos(θ)+1=1

3 cos (θ) + 1 = 1

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

Degrees

θ = 9 0^{\circ} + 36 0^{\circ} n, θ = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

3 cos (θ) + 1 = 1

Subtract

1

from both sides

3 cos (θ) + 1 - 1 = 1 - 1

Simplify

3 cos (θ) = 0

Divide both sides by

3

3 cos (θ) = 0

Divide both sides by

3

\frac{3 cos ( θ )}{3} = \frac{0}{3}

Simplify

cos (θ) = 0

cos (θ) = 0

General solutions for

cos (θ) = 0

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

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

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

sin (θ) = 3

2 cos (x) csc (x) = 2 cos (x)

- 2 cos^{2} (x) = 3 cos (x) + 1

1 + sec (x) = 3

tan (\frac{θ}{2} + \frac{π}{4}) = 1

What is the general solution for 3cos(θ)+1=1 ?
The general solution for 3cos(θ)+1=1 is θ= pi/2+2pin,θ=(3pi)/2+2pin