What is the general solution for 10cos(θ)=-5cos(θ) ?

The general solution for 10cos(θ)=-5cos(θ) is θ= pi/2+2pin,θ=(3pi)/2+2pin

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10cos(θ)=-5cos(θ)

10 cos (θ) = - 5 cos (θ)

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

10 cos (θ) = - 5 cos (θ)

Solve by substitution

10 cos (θ) = - 5 cos (θ)

Let:

cos (θ) = u

10 u = - 5 u

10 u = - 5 u : u = 0

10 u = - 5 u

Move

5 u

to the left side

10 u = - 5 u

Add

5 u

to both sides

10 u + 5 u = - 5 u + 5 u

Simplify

15 u = 0

15 u = 0

Divide both sides by

15

15 u = 0

Divide both sides by

15

\frac{15 u}{15} = \frac{0}{15}

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

tan (2 θ) = 1.33

2 cos (2 t) + 2 cos^{2} (t) = 0

5 \cdot cos (x) + 4 - 2 \cdot sin (x) = 0

sin (x) = - 0.54

tan (2 θ) = 1.82

What is the general solution for 10cos(θ)=-5cos(θ) ?
The general solution for 10cos(θ)=-5cos(θ) is θ= pi/2+2pin,θ=(3pi)/2+2pin