What is the general solution for 14cos(θ)-5=5cos(θ)-5 ?

The general solution for 14cos(θ)-5=5cos(θ)-5 is θ= pi/2+2pin,θ=(3pi)/2+2pin

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14cos(θ)-5=5cos(θ)-5

14 cos (θ) - 5 = 5 cos (θ) - 5

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

Degrees

θ = 9 0^{\circ} + 36 0^{\circ} n, θ = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

14 cos (θ) - 5 = 5 cos (θ) - 5

Solve by substitution

14 cos (θ) - 5 = 5 cos (θ) - 5

Let:

cos (θ) = u

14 u - 5 = 5 u - 5

14 u - 5 = 5 u - 5 : u = 0

14 u - 5 = 5 u - 5

Add

5

to both sides

14 u - 5 + 5 = 5 u - 5 + 5

Simplify

14 u = 5 u

Move

5 u

to the left side

14 u = 5 u

Subtract

5 u

from both sides

14 u - 5 u = 5 u - 5 u

Simplify

9 u = 0

9 u = 0

Divide both sides by

9

9 u = 0

Divide both sides by

9

\frac{9 u}{9} = \frac{0}{9}

Simplify

u = 0

u = 0

Substitute back

u = cos (θ)

cos (θ) = 0

cos (θ) = 0

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

cos (θ) = 0

General solutions for

cos (θ) = 0

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}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

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

Combine all the solutions

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

- 4 sin (2 θ) = 5 cos (2 θ)

tan (A) = \frac{3}{4}

2 sin (3 x) + 1 = 0

3 sin (θ) = cos (θ), 0 \leq θ \leq 2 π

cos (2 θ) = - \frac{2}{2}

What is the general solution for 14cos(θ)-5=5cos(θ)-5 ?
The general solution for 14cos(θ)-5=5cos(θ)-5 is θ= pi/2+2pin,θ=(3pi)/2+2pin