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

The general solution for -2cos(θ)-sqrt(3)=0 is θ=(5pi)/6+2pin,θ=(7pi)/6+2pin

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-2cos(θ)-sqrt(3)=0

- 2 cos (θ) - 3 = 0

θ = \frac{5 π}{6} + 2 πn, θ = \frac{7 π}{6} + 2 πn

Degrees

θ = 15 0^{\circ} + 36 0^{\circ} n, θ = 21 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{5 π}{6} + 2 πn, θ = \frac{7 π}{6} + 2 πn

θ = \frac{5 π}{6} + 2 πn, θ = \frac{7 π}{6} + 2 πn

4 cos^{2} (2 x) = 1

sec (x) = - 4.5973

2 cos (9 x) = 2 - 2 cos (9 x)

sin (2 x) = 32, - π \leq x \leq π

5 cos^{2} (x) - 2 = 0

What is the general solution for -2cos(θ)-sqrt(3)=0 ?
The general solution for -2cos(θ)-sqrt(3)=0 is θ=(5pi)/6+2pin,θ=(7pi)/6+2pin