What is the general solution for 6cos(θ)-3=0 ?

The general solution for 6cos(θ)-3=0 is θ= pi/3+2pin,θ=(5pi)/3+2pin

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6cos(θ)-3=0

6 cos (θ) - 3 = 0

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

Degrees

θ = 6 0^{\circ} + 36 0^{\circ} n, θ = 30 0^{\circ} + 36 0^{\circ} n

Solution steps

6 cos (θ) - 3 = 0

Move

3

to the right side

6 cos (θ) - 3 = 0

Add

3

to both sides

6 cos (θ) - 3 + 3 = 0 + 3

Simplify

6 cos (θ) = 3

6 cos (θ) = 3

Divide both sides by

6

6 cos (θ) = 3

Divide both sides by

6

\frac{6 cos ( θ )}{6} = \frac{3}{6}

Simplify

cos (θ) = \frac{1}{2}

cos (θ) = \frac{1}{2}

General solutions for

cos (θ) = \frac{1}{2}

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}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

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

3 cos (θ) + 3 = 2 sin^{2} (θ)

- 33 + 7 csc (\frac{θ}{5} + 2 3^{\circ}) = 53

10 sin (x) tan (x) + tan (x) = 0

2 sin^{2} (θ) + sin (θ) - 1 = 0, 0 \leq θ \leq 2 π

10 sin (x) cos (x) - 7 cos (x) = 0

What is the general solution for 6cos(θ)-3=0 ?
The general solution for 6cos(θ)-3=0 is θ= pi/3+2pin,θ=(5pi)/3+2pin