What is the general solution for (cos(3θ))(cos(θ))-1=(sin(3θ))(sin(θ)) ?

The general solution for (cos(3θ))(cos(θ))-1=(sin(3θ))(sin(θ)) is θ=(pin)/2

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(cos(3θ))(cos(θ))-1=(sin(3θ))(sin(θ))

(cos (3 θ)) (cos (θ)) - 1 = (sin (3 θ)) (sin (θ))

θ = \frac{πn}{2}

Degrees

θ = 0^{\circ} + 9 0^{\circ} n

Solution steps

(cos (3 θ)) (cos (θ)) - 1 = (sin (3 θ)) (sin (θ))

Subtract

sin (3 θ) sin (θ)

from both sides

cos (3 θ) cos (θ) - 1 - sin (3 θ) sin (θ) = 0

Rewrite using trig identities

cos (3 θ) cos (θ) - 1 - sin (3 θ) sin (θ)

Use the Angle Sum identity:

cos (s) cos (t) - sin (s) sin (t) = cos (s + t)

= - 1 + cos (3 θ + θ)

- 1 + cos (3 θ + θ) = 0

Move

1

to the right side

- 1 + cos (3 θ + θ) = 0

Add

1

to both sides

- 1 + cos (3 θ + θ) + 1 = 0 + 1

Simplify

cos (3 θ + θ) = 1

cos (3 θ + θ) = 1

General solutions for

cos (3 θ + θ) = 1

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}

3 θ + θ = 0 + 2 πn

3 θ + θ = 0 + 2 πn

Solve

3 θ + θ = 0 + 2 πn : θ = \frac{πn}{2}

3 θ + θ = 0 + 2 πn

Add similar elements:

3 θ + θ = 4 θ

4 θ = 0 + 2 πn

0 + 2 πn = 2 πn

4 θ = 2 πn

Divide both sides by

4

4 θ = 2 πn

Divide both sides by

4

\frac{4 θ}{4} = \frac{2 πn}{4}

Simplify

θ = \frac{πn}{2}

θ = \frac{πn}{2}

θ = \frac{πn}{2}

What is the general solution for (cos(3θ))(cos(θ))-1=(sin(3θ))(sin(θ)) ?
The general solution for (cos(3θ))(cos(θ))-1=(sin(3θ))(sin(θ)) is θ=(pin)/2