What is the general solution for cos(θ)cos(2θ)+sin(θ)sin(2θ)=(sqrt(2))/2 ?

The general solution for cos(θ)cos(2θ)+sin(θ)sin(2θ)=(sqrt(2))/2 is θ=-pi/4-2pin,θ=-(7pi)/4-2pin

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cos(θ)cos(2θ)+sin(θ)sin(2θ)=(sqrt(2))/2

Solution

cos (θ) cos (2 θ) + sin (θ) sin (2 θ) = \frac{2}{2}

Solution

θ = - \frac{π}{4} - 2 πn, θ = - \frac{7 π}{4} - 2 πn

Degrees

θ = - 4 5^{\circ} - 36 0^{\circ} n, θ = - 31 5^{\circ} - 36 0^{\circ} n

Solution steps

cos (θ) cos (2 θ) + sin (θ) sin (2 θ) = \frac{2}{2}

Rewrite using trig identities

cos (θ) cos (2 θ) + sin (θ) sin (2 θ)

Use the Angle Difference identity:

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

= cos (θ - 2 θ)

cos (θ - 2 θ) = \frac{2}{2}

General solutions for

cos (θ - 2 θ) = \frac{2}{2}

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}

θ - 2 θ = \frac{π}{4} + 2 πn, θ - 2 θ = \frac{7 π}{4} + 2 πn

θ - 2 θ = \frac{π}{4} + 2 πn, θ - 2 θ = \frac{7 π}{4} + 2 πn

Solve

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

θ - 2 θ = \frac{π}{4} + 2 πn

Add similar elements:

θ - 2 θ = - θ

- θ = \frac{π}{4} + 2 πn

Divide both sides by

- 1

- θ = \frac{π}{4} + 2 πn

Divide both sides by

- 1

\frac{- θ}{- 1} = \frac{\frac{π}{4}}{- 1} + \frac{2 πn}{- 1}

Simplify

\frac{- θ}{- 1} = \frac{\frac{π}{4}}{- 1} + \frac{2 πn}{- 1}

Simplify

\frac{- θ}{- 1} : θ

\frac{- θ}{- 1}

Apply the fraction rule:

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

= \frac{θ}{1}

Apply rule

\frac{a}{1} = a

= θ

Simplify

\frac{\frac{π}{4}}{- 1} + \frac{2 πn}{- 1} : - \frac{π}{4} - 2 πn

\frac{\frac{π}{4}}{- 1} + \frac{2 πn}{- 1}

\frac{\frac{π}{4}}{- 1} = - \frac{π}{4}

\frac{\frac{π}{4}}{- 1}

Apply the fraction rule:

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

= - \frac{\frac{π}{4}}{1}

Apply the fraction rule:

\frac{a}{1} = a

\frac{\frac{π}{4}}{1} = \frac{π}{4}

= - \frac{π}{4}

= - \frac{π}{4} + \frac{2 πn}{- 1}

\frac{2 πn}{- 1} = - 2 πn

\frac{2 πn}{- 1}

Apply the fraction rule:

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

= - \frac{2 πn}{1}

Apply rule

\frac{a}{1} = a

= - 2 πn

= - \frac{π}{4} - 2 πn

θ = - \frac{π}{4} - 2 πn

θ = - \frac{π}{4} - 2 πn

θ = - \frac{π}{4} - 2 πn

Solve

θ - 2 θ = \frac{7 π}{4} + 2 πn : θ = - \frac{7 π}{4} - 2 πn

θ - 2 θ = \frac{7 π}{4} + 2 πn

Add similar elements:

θ - 2 θ = - θ

- θ = \frac{7 π}{4} + 2 πn

Divide both sides by

- 1

- θ = \frac{7 π}{4} + 2 πn

Divide both sides by

- 1

\frac{- θ}{- 1} = \frac{\frac{7 π}{4}}{- 1} + \frac{2 πn}{- 1}

Simplify

\frac{- θ}{- 1} = \frac{\frac{7 π}{4}}{- 1} + \frac{2 πn}{- 1}

Simplify

\frac{- θ}{- 1} : θ

\frac{- θ}{- 1}

Apply the fraction rule:

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

= \frac{θ}{1}

Apply rule

\frac{a}{1} = a

= θ

Simplify

\frac{\frac{7 π}{4}}{- 1} + \frac{2 πn}{- 1} : - \frac{7 π}{4} - 2 πn

\frac{\frac{7 π}{4}}{- 1} + \frac{2 πn}{- 1}

\frac{\frac{7 π}{4}}{- 1} = - \frac{7 π}{4}

\frac{\frac{7 π}{4}}{- 1}

Apply the fraction rule:

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

= - \frac{\frac{7 π}{4}}{1}

Apply the fraction rule:

\frac{a}{1} = a

\frac{\frac{7 π}{4}}{1} = \frac{7 π}{4}

= - \frac{7 π}{4}

= - \frac{7 π}{4} + \frac{2 πn}{- 1}

\frac{2 πn}{- 1} = - 2 πn

\frac{2 πn}{- 1}

Apply the fraction rule:

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

= - \frac{2 πn}{1}

Apply rule

\frac{a}{1} = a

= - 2 πn

= - \frac{7 π}{4} - 2 πn

θ = - \frac{7 π}{4} - 2 πn

θ = - \frac{7 π}{4} - 2 πn

θ = - \frac{7 π}{4} - 2 πn

θ = - \frac{π}{4} - 2 πn, θ = - \frac{7 π}{4} - 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for cos(θ)cos(2θ)+sin(θ)sin(2θ)=(sqrt(2))/2 ?
The general solution for cos(θ)cos(2θ)+sin(θ)sin(2θ)=(sqrt(2))/2 is θ=-pi/4-2pin,θ=-(7pi)/4-2pin