What is the general solution for cos(4θ)=cos(2θ),180<= θ<= 360 ?

Q: What is the general solution for cos(4θ)=cos(2θ),180<= θ<= 360 ?

The general solution for cos(4θ)=cos(2θ),180<= θ<= 360 is θ=180,θ=240,θ=300,θ=360

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cos(4θ)=cos(2θ),180<= θ<= 360

Solution

cos (4 θ) = cos (2 θ), 18 0^{\circ} \leq θ \leq 36 0^{\circ}

Solution

θ = 18 0^{\circ}, θ = 24 0^{\circ}, θ = 30 0^{\circ}, θ = 36 0^{\circ}

Radians

θ = π, θ = \frac{4 π}{3}, θ = \frac{5 π}{3}, θ = 2 π

Solution steps

cos (4 θ) = cos (2 θ), 18 0^{\circ} \leq θ \leq 36 0^{\circ}

Subtract

cos (2 θ)

from both sides

cos (4 θ) - cos (2 θ) = 0

Rewrite using trig identities

- cos (2 θ) + cos (4 θ)

Use the Sum to Product identity:

cos (s) - cos (t) = - 2 sin (\frac{s + t}{2}) sin (\frac{s - t}{2})

= - 2 sin (\frac{4 θ + 2 θ}{2}) sin (\frac{4 θ - 2 θ}{2})

Simplify

- 2 sin (\frac{4 θ + 2 θ}{2}) sin (\frac{4 θ - 2 θ}{2}) : - 2 sin (3 θ) sin (θ)

- 2 sin (\frac{4 θ + 2 θ}{2}) sin (\frac{4 θ - 2 θ}{2})

\frac{4 θ + 2 θ}{2} = 3 θ

\frac{4 θ + 2 θ}{2}

Add similar elements:

4 θ + 2 θ = 6 θ

= \frac{6 θ}{2}

Divide the numbers:

\frac{6}{2} = 3

= 3 θ

= - 2 sin (3 θ) sin (\frac{4 θ - 2 θ}{2})

\frac{4 θ - 2 θ}{2} = θ

\frac{4 θ - 2 θ}{2}

Add similar elements:

4 θ - 2 θ = 2 θ

= \frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

= - 2 sin (3 θ) sin (θ)

= - 2 sin (3 θ) sin (θ)

- 2 sin (3 θ) sin (θ) = 0

Solving each part separately

sin (3 θ) = 0 or sin (θ) = 0

sin (3 θ) = 0, 18 0^{\circ} \leq θ \leq 36 0^{\circ} : θ = 18 0^{\circ}, θ = 24 0^{\circ}, θ = 30 0^{\circ}, θ = 36 0^{\circ}

sin (3 θ) = 0, 18 0^{\circ} \leq θ \leq 36 0^{\circ}

General solutions for

sin (3 θ) = 0

sin (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

3 θ = 0 + 36 0^{\circ} n, 3 θ = 18 0^{\circ} + 36 0^{\circ} n

3 θ = 0 + 36 0^{\circ} n, 3 θ = 18 0^{\circ} + 36 0^{\circ} n

Solve

3 θ = 0 + 36 0^{\circ} n : θ = \frac{36 0 ^{\circ} n}{3}

3 θ = 0 + 36 0^{\circ} n

0 + 36 0^{\circ} n = 36 0^{\circ} n

3 θ = 36 0^{\circ} n

Divide both sides by

3

3 θ = 36 0^{\circ} n

Divide both sides by

3

\frac{3 θ}{3} = \frac{36 0 ^{\circ} n}{3}

Simplify

θ = \frac{36 0 ^{\circ} n}{3}

θ = \frac{36 0 ^{\circ} n}{3}

Solve

3 θ = 18 0^{\circ} + 36 0^{\circ} n : θ = 6 0^{\circ} + \frac{36 0 ^{\circ} n}{3}

3 θ = 18 0^{\circ} + 36 0^{\circ} n

Divide both sides by

3

3 θ = 18 0^{\circ} + 36 0^{\circ} n

Divide both sides by

3

\frac{3 θ}{3} = 6 0^{\circ} + \frac{36 0 ^{\circ} n}{3}

Simplify

θ = 6 0^{\circ} + \frac{36 0 ^{\circ} n}{3}

θ = 6 0^{\circ} + \frac{36 0 ^{\circ} n}{3}

θ = \frac{36 0 ^{\circ} n}{3}, θ = 6 0^{\circ} + \frac{36 0 ^{\circ} n}{3}

Solutions for the range

18 0^{\circ} \leq θ \leq 36 0^{\circ}

θ = 18 0^{\circ}, θ = 24 0^{\circ}, θ = 30 0^{\circ}, θ = 36 0^{\circ}

sin (θ) = 0, 18 0^{\circ} \leq θ \leq 36 0^{\circ} : θ = 18 0^{\circ}, θ = 36 0^{\circ}

sin (θ) = 0, 18 0^{\circ} \leq θ \leq 36 0^{\circ}

General solutions for

sin (θ) = 0

sin (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

θ = 0 + 36 0^{\circ} n, θ = 18 0^{\circ} + 36 0^{\circ} n

θ = 0 + 36 0^{\circ} n, θ = 18 0^{\circ} + 36 0^{\circ} n

Solve

θ = 0 + 36 0^{\circ} n : θ = 36 0^{\circ} n

θ = 0 + 36 0^{\circ} n

0 + 36 0^{\circ} n = 36 0^{\circ} n

θ = 36 0^{\circ} n

θ = 36 0^{\circ} n, θ = 18 0^{\circ} + 36 0^{\circ} n

Solutions for the range

18 0^{\circ} \leq θ \leq 36 0^{\circ}

θ = 18 0^{\circ}, θ = 36 0^{\circ}

Combine all the solutions

θ = 18 0^{\circ}, θ = 24 0^{\circ}, θ = 30 0^{\circ}, θ = 36 0^{\circ}

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8cos(2x)=8-8cos(2x)

8 cos (2 x) = 8 - 8 cos (2 x)

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- (2) cos (10 t) = (6) sin (10 t)

cot(x)= 1/4

cot (x) = \frac{1}{4}

tan(x)=0,0<x<2pi

tan (x) = 0, 0 < x < 2 π

Frequently Asked Questions (FAQ)

What is the general solution for cos(4θ)=cos(2θ),180<= θ<= 360 ?
The general solution for cos(4θ)=cos(2θ),180<= θ<= 360 is θ=180,θ=240,θ=300,θ=360