What is the general solution for cos(θ)tan^2(θ)=3cos(θ) ?

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

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cos(θ)tan^2(θ)=3cos(θ)

Solution

cos (θ) tan^{2} (θ) = 3 cos (θ)

Solution

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

Degrees

θ = 12 0^{\circ} + 18 0^{\circ} n, θ = 6 0^{\circ} + 18 0^{\circ} n

Solution steps

cos (θ) tan^{2} (θ) = 3 cos (θ)

Subtract

3 cos (θ)

from both sides

tan^{2} (θ) cos (θ) - 3 cos (θ) = 0

Factor

tan^{2} (θ) cos (θ) - 3 cos (θ) : cos (θ) (tan (θ) + 3) (tan (θ) - 3)

tan^{2} (θ) cos (θ) - 3 cos (θ)

Factor out common term

cos (θ)

= cos (θ) (tan^{2} (θ) - 3)

Factor

tan^{2} (θ) - 3 : (tan (θ) + 3) (tan (θ) - 3)

tan^{2} (θ) - 3

Apply radical rule:

a = (a)^{2}

3 = (3)^{2}

= tan^{2} (θ) - (3)^{2}

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

tan^{2} (θ) - (3)^{2} = (tan (θ) + 3) (tan (θ) - 3)

= (tan (θ) + 3) (tan (θ) - 3)

= cos (θ) (tan (θ) + 3) (tan (θ) - 3)

cos (θ) (tan (θ) + 3) (tan (θ) - 3) = 0

Solving each part separately

cos (θ) = 0 or tan (θ) + 3 = 0 or tan (θ) - 3 = 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

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

tan (θ) + 3 = 0

Move

3

to the right side

tan (θ) + 3 = 0

Subtract

3

from both sides

tan (θ) + 3 - 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

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

tan (θ) - 3 = 0

Move

3

to the right side

tan (θ) - 3 = 0

Add

3

to both sides

tan (θ) - 3 + 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{π}{3} + πn

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

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5cos(x)+1=1

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cosh (z) = 1

Frequently Asked Questions (FAQ)

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