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= pi/4+(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{π}{4} + \frac{πn}{2}

Degrees

θ = 4 5^{\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

Subtract

1

from 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 = π + 2 πn

3 θ + θ = π + 2 πn

Solve

3 θ + θ = π + 2 πn : θ = \frac{π}{4} + \frac{πn}{2}

3 θ + θ = π + 2 πn

Add similar elements:

3 θ + θ = 4 θ

4 θ = π + 2 πn

Divide both sides by

4

4 θ = π + 2 πn

Divide both sides by

4

\frac{4 θ}{4} = \frac{π}{4} + \frac{2 πn}{4}

Simplify

θ = \frac{π}{4} + \frac{πn}{2}

θ = \frac{π}{4} + \frac{πn}{2}

θ = \frac{π}{4} + \frac{πn}{2}

4 sin (x) - 3 = 0

9 sin (x) tan (x) = 2 tan (x)

- 5 sin (2 x) = 0

- 2 sin^{2} (x) = - sin (x) - 1

tan (x) = \frac{5}{7}

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= pi/4+(pin)/2