What is the general solution for tan(θ)(3cos^2(θ)-sin^2(θ))=0 ?

The general solution for tan(θ)(3cos^2(θ)-sin^2(θ))=0 is θ=pin,θ=(2pi)/3+pin,θ= pi/3+pin

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tan(θ)(3cos^2(θ)-sin^2(θ))=0

Solution

tan (θ) (3 cos^{2} (θ) - sin^{2} (θ)) = 0

Solution

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

Degrees

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

Solution steps

tan (θ) (3 cos^{2} (θ) - sin^{2} (θ)) = 0

Solving each part separately

tan (θ) = 0 or 3 cos^{2} (θ) - sin^{2} (θ) = 0

tan (θ) = 0 : θ = πn

tan (θ) = 0

General solutions for

tan (θ) = 0

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 = 0 + πn

θ = 0 + πn

Solve

θ = 0 + πn : θ = πn

θ = 0 + πn

0 + πn = πn

θ = πn

θ = πn

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

3 cos^{2} (θ) - sin^{2} (θ) = 0

Factor

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

3 cos^{2} (θ) - sin^{2} (θ)

Rewrite

3 cos^{2} (θ) - sin^{2} (θ)

(3 cos (θ))^{2} - sin^{2} (θ)

3 cos^{2} (θ) - sin^{2} (θ)

Apply radical rule:

a = (a)^{2}

3 = (3)^{2}

= (3)^{2} cos^{2} (θ) - sin^{2} (θ)

Apply exponent rule:

a^{m} b^{m} = (ab)^{m}

(3)^{2} cos^{2} (θ) = (3 cos (θ))^{2}

= (3 cos (θ))^{2} - sin^{2} (θ)

= (3 cos (θ))^{2} - sin^{2} (θ)

Apply Difference of Two Squares Formula:

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

(3 cos (θ))^{2} - sin^{2} (θ) = (3 cos (θ) + sin (θ)) (3 cos (θ) - sin (θ))

= (3 cos (θ) + sin (θ)) (3 cos (θ) - sin (θ))

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

Solving each part separately

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

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

3 cos (θ) + sin (θ) = 0

Rewrite using trig identities

3 cos (θ) + sin (θ) = 0

Divide both sides by

cos (θ), cos (θ) \neq = 0

\frac{3 cos ( θ ) + sin ( θ )}{cos ( θ )} = \frac{0}{cos ( θ )}

Simplify

3 + \frac{sin ( θ )}{cos ( θ )} = 0

Use the basic trigonometric identity:

\frac{sin ( x )}{cos ( x )} = tan (x)

3 + tan (θ) = 0

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

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

3 cos (θ) - sin (θ) = 0

Rewrite using trig identities

3 cos (θ) - sin (θ) = 0

Divide both sides by

cos (θ), cos (θ) \neq = 0

\frac{3 cos ( θ ) - sin ( θ )}{cos ( θ )} = \frac{0}{cos ( θ )}

Simplify

3 - \frac{sin ( θ )}{cos ( θ )} = 0

Use the basic trigonometric identity:

\frac{sin ( x )}{cos ( x )} = tan (x)

3 - tan (θ) = 0

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

Divide both sides by

- 1

- tan (θ) = - 3

Divide both sides by

- 1

\frac{- tan ( θ )}{- 1} = \frac{- 3}{- 1}

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{2 π}{3} + πn, θ = \frac{π}{3} + πn

Combine all the solutions

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

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2 cos^{2} (w) + 5 cos (w) + 3 = 0

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tan (x) = - 135

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3 cos (x) - 4 sin (x) = - 3

Frequently Asked Questions (FAQ)

What is the general solution for tan(θ)(3cos^2(θ)-sin^2(θ))=0 ?
The general solution for tan(θ)(3cos^2(θ)-sin^2(θ))=0 is θ=pin,θ=(2pi)/3+pin,θ= pi/3+pin