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

The general solution for sqrt(3)cos(θ)=-cos(θ)tan(θ) is θ=(2pi)/3+pin

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sqrt(3)cos(θ)=-cos(θ)tan(θ)

3 cos (θ) = - cos (θ) tan (θ)

θ = \frac{2 π}{3} + πn

Degrees

θ = 12 0^{\circ} + 18 0^{\circ} n

Solution steps

3 cos (θ) = - cos (θ) tan (θ)

Subtract

- cos (θ) tan (θ)

from both sides

3 cos (θ) + cos (θ) tan (θ) = 0

Factor

3 cos (θ) + cos (θ) tan (θ) : cos (θ) (3 + tan (θ))

3 cos (θ) + cos (θ) tan (θ)

Factor out common term

cos (θ)

= cos (θ) (3 + tan (θ))

cos (θ) (3 + tan (θ)) = 0

Solving each part separately

cos (θ) = 0 or 3 + tan (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

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

3 + tan (θ) = 0

Move

3

to the right side

3 + tan (θ) = 0

Subtract

3

from both sides

3 + tan (θ) - 3 = 0 - 3

Simplify

tan (θ) = - 3

tan (θ) = - 3

General solutions for

tan (θ) = - 3

tan (x)

periodicity table with

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

θ = \frac{2 π}{3} + πn

θ = \frac{2 π}{3} + πn

Combine all the solutions

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

Since the equation is undefined for:

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

θ = \frac{2 π}{3} + πn

cos (2 x) = cos (x) - cos (3 0^{\circ}) [180]

1 - 2 tan (x) = tan^{2} (x)

- 13 a \cdot sin (2 x) - 13 b \cdot cos (2 x) = 0

3 + sin (x) = 2

1 - sin (A) cos (A) = (sin (A) - cos (A))^{2}

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