What is the general solution for 4cos(2θ)=8sin^2(θ) ?

The general solution for 4cos(2θ)=8sin^2(θ) is θ=(5pi)/6+pin,θ= pi/6+pin

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4cos(2θ)=8sin^2(θ)

Solution

4 cos (2 θ) = 8 sin^{2} (θ)

Solution

θ = \frac{5 π}{6} + πn, θ = \frac{π}{6} + πn

Degrees

θ = 15 0^{\circ} + 18 0^{\circ} n, θ = 3 0^{\circ} + 18 0^{\circ} n

Solution steps

4 cos (2 θ) = 8 sin^{2} (θ)

Subtract

8 sin^{2} (θ)

from both sides

4 cos (2 θ) - 8 sin^{2} (θ) = 0

Rewrite using trig identities

4 cos (2 θ) - 8 sin^{2} (θ)

Use the Double Angle identity:

cos (2 x) = cos^{2} (x) - sin^{2} (x)

= 4 (cos^{2} (θ) - sin^{2} (θ)) - 8 sin^{2} (θ)

Simplify

4 (cos^{2} (θ) - sin^{2} (θ)) - 8 sin^{2} (θ) : 4 cos^{2} (θ) - 12 sin^{2} (θ)

4 (cos^{2} (θ) - sin^{2} (θ)) - 8 sin^{2} (θ)

Expand

4 (cos^{2} (θ) - sin^{2} (θ)) : 4 cos^{2} (θ) - 4 sin^{2} (θ)

4 (cos^{2} (θ) - sin^{2} (θ))

Apply the distributive law:

a (b - c) = ab - a c

a = 4, b = cos^{2} (θ), c = sin^{2} (θ)

= 4 cos^{2} (θ) - 4 sin^{2} (θ)

= 4 cos^{2} (θ) - 4 sin^{2} (θ) - 8 sin^{2} (θ)

Add similar elements:

- 4 sin^{2} (θ) - 8 sin^{2} (θ) = - 12 sin^{2} (θ)

= 4 cos^{2} (θ) - 12 sin^{2} (θ)

= 4 cos^{2} (θ) - 12 sin^{2} (θ)

- 12 sin^{2} (θ) + 4 cos^{2} (θ) = 0

Factor

- 12 sin^{2} (θ) + 4 cos^{2} (θ) : 4 (cos (θ) + 3 sin (θ)) (cos (θ) - 3 sin (θ))

- 12 sin^{2} (θ) + 4 cos^{2} (θ)

Rewrite

- 12

3 \cdot 4

= 3 \cdot 4 sin^{2} (θ) + 4 cos^{2} (θ)

Factor out common term

4

= 4 (- 3 sin^{2} (θ) + cos^{2} (θ))

Factor

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

cos^{2} (θ) - 3 sin^{2} (θ)

Rewrite

cos^{2} (θ) - 3 sin^{2} (θ)

cos^{2} (θ) - (3 sin (θ))^{2}

cos^{2} (θ) - 3 sin^{2} (θ)

Apply radical rule:

a = (a)^{2}

3 = (3)^{2}

= cos^{2} (θ) - (3)^{2} sin^{2} (θ)

Apply exponent rule:

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

(3)^{2} sin^{2} (θ) = (3 sin (θ))^{2}

= cos^{2} (θ) - (3 sin (θ))^{2}

= cos^{2} (θ) - (3 sin (θ))^{2}

Apply Difference of Two Squares Formula:

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

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

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

= 4 (cos (θ) + 3 sin (θ)) (cos (θ) - 3 sin (θ))

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

Solving each part separately

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

cos (θ) + 3 sin (θ) = 0 : θ = \frac{5 π}{6} + πn

cos (θ) + 3 sin (θ) = 0

Rewrite using trig identities

cos (θ) + 3 sin (θ) = 0

Divide both sides by

cos (θ), cos (θ) \neq = 0

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

Simplify

1 + \frac{3 sin ( θ )}{cos ( θ )} = 0

Use the basic trigonometric identity:

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

1 + 3 tan (θ) = 0

1 + 3 tan (θ) = 0

Move

1

to the right side

1 + 3 tan (θ) = 0

Subtract

1

from both sides

1 + 3 tan (θ) - 1 = 0 - 1

Simplify

3 tan (θ) = - 1

3 tan (θ) = - 1

Divide both sides by

3

3 tan (θ) = - 1

Divide both sides by

3

\frac{3 tan ( θ )}{3} = \frac{- 1}{3}

Simplify

\frac{3 tan ( θ )}{3} = \frac{- 1}{3}

Simplify

\frac{3 tan ( θ )}{3} : tan (θ)

\frac{3 tan ( θ )}{3}

Cancel the common factor:

3

= tan (θ)

Simplify

\frac{- 1}{3} : - \frac{3}{3}

\frac{- 1}{3}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{1}{3}

Rationalize

- \frac{1}{3} : - \frac{3}{3}

- \frac{1}{3}

Multiply by the conjugate

\frac{3}{3}

= - \frac{1 \cdot 3}{3 3}

1 \cdot 3 = 3

33 = 3

33

Apply radical rule:

a a = a

33 = 3

= 3

= - \frac{3}{3}

= - \frac{3}{3}

tan (θ) = - \frac{3}{3}

tan (θ) = - \frac{3}{3}

tan (θ) = - \frac{3}{3}

General solutions for

tan (θ) = - \frac{3}{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{5 π}{6} + πn

θ = \frac{5 π}{6} + πn

cos (θ) - 3 sin (θ) = 0 : θ = \frac{π}{6} + πn

cos (θ) - 3 sin (θ) = 0

Rewrite using trig identities

cos (θ) - 3 sin (θ) = 0

Divide both sides by

cos (θ), cos (θ) \neq = 0

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

Simplify

1 - \frac{3 sin ( θ )}{cos ( θ )} = 0

Use the basic trigonometric identity:

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

1 - 3 tan (θ) = 0

1 - 3 tan (θ) = 0

Move

1

to the right side

1 - 3 tan (θ) = 0

Subtract

1

from both sides

1 - 3 tan (θ) - 1 = 0 - 1

Simplify

- 3 tan (θ) = - 1

- 3 tan (θ) = - 1

Divide both sides by

- 3

- 3 tan (θ) = - 1

Divide both sides by

- 3

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

Simplify

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

Simplify

\frac{- 3 tan ( θ )}{- 3} : tan (θ)

\frac{- 3 tan ( θ )}{- 3}

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

= \frac{3 tan ( θ )}{3}

Cancel the common factor:

3

= tan (θ)

Simplify

\frac{- 1}{- 3} : \frac{3}{3}

\frac{- 1}{- 3}

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

= \frac{1}{3}

Rationalize

\frac{1}{3} : \frac{3}{3}

\frac{1}{3}

Multiply by the conjugate

\frac{3}{3}

= \frac{1 \cdot 3}{3 3}

1 \cdot 3 = 3

33 = 3

33

Apply radical rule:

a a = a

33 = 3

= 3

= \frac{3}{3}

= \frac{3}{3}

tan (θ) = \frac{3}{3}

tan (θ) = \frac{3}{3}

tan (θ) = \frac{3}{3}

General solutions for

tan (θ) = \frac{3}{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}{6} + πn

θ = \frac{π}{6} + πn

Combine all the solutions

θ = \frac{5 π}{6} + πn, θ = \frac{π}{6} + πn

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Frequently Asked Questions (FAQ)

What is the general solution for 4cos(2θ)=8sin^2(θ) ?
The general solution for 4cos(2θ)=8sin^2(θ) is θ=(5pi)/6+pin,θ= pi/6+pin