What is the general solution for 7cos(θ)=sqrt(2)+9cos(θ) ?

The general solution for 7cos(θ)=sqrt(2)+9cos(θ) is θ=(3pi)/4+2pin,θ=(5pi)/4+2pin

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7cos(θ)=sqrt(2)+9cos(θ)

7 cos (θ) = 2 + 9 cos (θ)

θ = \frac{3 π}{4} + 2 πn, θ = \frac{5 π}{4} + 2 πn

Degrees

θ = 13 5^{\circ} + 36 0^{\circ} n, θ = 22 5^{\circ} + 36 0^{\circ} n

Solution steps

7 cos (θ) = 2 + 9 cos (θ)

Solve by substitution

7 cos (θ) = 2 + 9 cos (θ)

Let:

cos (θ) = u

7 u = 2 + 9 u

7 u = 2 + 9 u : u = - \frac{2}{2}

7 u = 2 + 9 u

Move

9 u

to the left side

7 u = 2 + 9 u

Subtract

9 u

from both sides

7 u - 9 u = 2 + 9 u - 9 u

Simplify

- 2 u = 2

- 2 u = 2

Divide both sides by

- 2

- 2 u = 2

Divide both sides by

- 2

\frac{- 2 u}{- 2} = \frac{2}{- 2}

Simplify

u = - \frac{2}{2}

u = - \frac{2}{2}

Substitute back

u = cos (θ)

cos (θ) = - \frac{2}{2}

cos (θ) = - \frac{2}{2}

cos (θ) = - \frac{2}{2} : θ = \frac{3 π}{4} + 2 πn, θ = \frac{5 π}{4} + 2 πn

cos (θ) = - \frac{2}{2}

General solutions for

cos (θ) = - \frac{2}{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{3 π}{4} + 2 πn, θ = \frac{5 π}{4} + 2 πn

θ = \frac{3 π}{4} + 2 πn, θ = \frac{5 π}{4} + 2 πn

Combine all the solutions

θ = \frac{3 π}{4} + 2 πn, θ = \frac{5 π}{4} + 2 πn

sec (x) \cdot cot (x) - 2 cot (x) + sec (x) = 2

3 sin^{2} (x) - 10 sin (x) - 8 = 0

2 \cdot 9.8 \cdot L (1 - cos (θ)) = 0.018

cos (7 x) = sin (4 x + 2)

12 cos (β) - 5 sin (β) = 4.7

What is the general solution for 7cos(θ)=sqrt(2)+9cos(θ) ?
The general solution for 7cos(θ)=sqrt(2)+9cos(θ) is θ=(3pi)/4+2pin,θ=(5pi)/4+2pin