What is the general solution for solvefor θ,2cos(θ)-sqrt(3)=0 ?

The general solution for solvefor θ,2cos(θ)-sqrt(3)=0 is θ= pi/6+2pin,θ=(11pi)/6+2pin

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solvefor θ,2cos(θ)-sqrt(3)=0

solve for

θ, 2 cos (θ) - 3 = 0

θ = \frac{π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

Degrees

θ = 3 0^{\circ} + 36 0^{\circ} n, θ = 33 0^{\circ} + 36 0^{\circ} n

Solution steps

2 cos (θ) - 3 = 0

Move

3

to the right side

2 cos (θ) - 3 = 0

Add

3

to both sides

2 cos (θ) - 3 + 3 = 0 + 3

Simplify

2 cos (θ) = 3

2 cos (θ) = 3

Divide both sides by

2

2 cos (θ) = 3

Divide both sides by

2

\frac{2 cos ( θ )}{2} = \frac{3}{2}

Simplify

cos (θ) = \frac{3}{2}

cos (θ) = \frac{3}{2}

General solutions for

cos (θ) = \frac{3}{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}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

θ = \frac{π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

3 cos (4 θ - 2 6^{\circ}) + 5 = 7, - 3 0^{\circ} \leq θ < 4 5^{\circ}

tan^{2} (θ) - tan (θ) = 6, 0^{\circ} \leq θ < 18 0^{\circ}

sin (x) = \frac{2 3}{3}

\frac{sin ( 2 x )}{cos ( x )} = 0

12 sin^{2} (x) = 5 cos (x) + 9

What is the general solution for solvefor θ,2cos(θ)-sqrt(3)=0 ?
The general solution for solvefor θ,2cos(θ)-sqrt(3)=0 is θ= pi/6+2pin,θ=(11pi)/6+2pin