What is the general solution for (2cos(θ)+1)(cos(θ)-2)=0 ?

The general solution for (2cos(θ)+1)(cos(θ)-2)=0 is θ=(2pi)/3+2pin,θ=(4pi)/3+2pin

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(2cos(θ)+1)(cos(θ)-2)=0

(2 cos (θ) + 1) (cos (θ) - 2) = 0

θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

Degrees

θ = 12 0^{\circ} + 36 0^{\circ} n, θ = 24 0^{\circ} + 36 0^{\circ} n

Solution steps

(2 cos (θ) + 1) (cos (θ) - 2) = 0

Solving each part separately

2 cos (θ) + 1 = 0 or cos (θ) - 2 = 0

2 cos (θ) + 1 = 0 : θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

2 cos (θ) + 1 = 0

Move

1

to the right side

2 cos (θ) + 1 = 0

Subtract

1

from both sides

2 cos (θ) + 1 - 1 = 0 - 1

Simplify

2 cos (θ) = - 1

2 cos (θ) = - 1

Divide both sides by

2

2 cos (θ) = - 1

Divide both sides by

2

\frac{2 cos ( θ )}{2} = \frac{- 1}{2}

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{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

cos (θ) - 2 = 0 :

No Solution

cos (θ) - 2 = 0

Move

2

to the right side

cos (θ) - 2 = 0

Add

2

to both sides

cos (θ) - 2 + 2 = 0 + 2

Simplify

cos (θ) = 2

cos (θ) = 2

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

4 cos (2 x) - 1 = 1

sin (2 x) = cot (x)

cos (θ) = 3

cos (θ) = 4

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

What is the general solution for (2cos(θ)+1)(cos(θ)-2)=0 ?
The general solution for (2cos(θ)+1)(cos(θ)-2)=0 is θ=(2pi)/3+2pin,θ=(4pi)/3+2pin