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

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

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2cos(θ)-sqrt(3)=0,0<= θ<= 2pi

2 cos (θ) - 3 = 0, 0 \leq θ \leq 2 π

θ = \frac{π}{6}, θ = \frac{11 π}{6}

Degrees

θ = 3 0^{\circ}, θ = 33 0^{\circ}

Solution steps

2 cos (θ) - 3 = 0, 0 \leq θ \leq 2 π

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

Solutions for the range

0 \leq θ \leq 2 π

θ = \frac{π}{6}, θ = \frac{11 π}{6}

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

3 sin (x) - 3 cos (x) = 3

tan (2 x) cos (2 x) + cot (2 x) sin (2 x) = 1

sin^{2} (2 x) - cos^{2} (2 x) = \frac{1}{2}

sin (x) = \frac{1.2}{10}

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