What is the general solution for cos(θ)cos(2θ)+sin(θ)sin(2θ)= 1/2 ?

The general solution for cos(θ)cos(2θ)+sin(θ)sin(2θ)= 1/2 is θ=-pi/3-2pin,θ=-(5pi)/3-2pin

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cos(θ)cos(2θ)+sin(θ)sin(2θ)= 1/2

cos (θ) cos (2 θ) + sin (θ) sin (2 θ) = \frac{1}{2}

θ = - \frac{π}{3} - 2 πn, θ = - \frac{5 π}{3} - 2 πn

Solution steps

cos (θ) cos (2 θ) + sin (θ) sin (2 θ) = \frac{1}{2}

Rewrite using trig identities

cos (θ - 2 θ) = \frac{1}{2}

General solutions for

cos (θ - 2 θ) = \frac{1}{2}

θ - 2 θ = \frac{π}{3} + 2 πn, θ - 2 θ = \frac{5 π}{3} + 2 πn

Solve

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

Solve

θ - 2 θ = \frac{5 π}{3} + 2 πn : θ = - \frac{5 π}{3} - 2 πn

θ = - \frac{π}{3} - 2 πn, θ = - \frac{5 π}{3} - 2 πn

8 sin (x) tan (x) + tan (x) = 0

sin (x) - 3 = cos (x) - 3

sin (x) + 4 csc (x) + 5 = 0

sin (x) = \frac{5}{9}

sin (x) = \frac{5}{3}

What is the general solution for cos(θ)cos(2θ)+sin(θ)sin(2θ)= 1/2 ?
The general solution for cos(θ)cos(2θ)+sin(θ)sin(2θ)= 1/2 is θ=-pi/3-2pin,θ=-(5pi)/3-2pin