What is the general solution for (18cos(2θ)+27cos(θ))/2 =0 ?

The general solution for (18cos(2θ)+27cos(θ))/2 =0 is θ=1.13140…+2pin,θ=2pi-1.13140…+2pin

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(18cos(2θ)+27cos(θ))/2 =0

\frac{18 cos ( 2 θ ) + 27 cos ( θ )}{2} = 0

θ = 1.13140 \dots + 2 πn, θ = 2 π - 1.13140 \dots + 2 πn

Solution steps

\frac{18 cos ( 2 θ ) + 27 cos ( θ )}{2} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

18 cos (2 θ) + 27 cos (θ) = 0

Rewrite using trig identities

(- 1 + 2 cos^{2} (θ)) \cdot 18 + 27 cos (θ) = 0

Solve by substitution

cos (θ) = \frac{- 3 + 41}{8}, cos (θ) = - \frac{3 + 41}{8}

cos (θ) = \frac{- 3 + 41}{8} : θ = arccos (\frac{- 3 + 41}{8}) + 2 πn, θ = 2 π - arccos (\frac{- 3 + 41}{8}) + 2 πn

cos (θ) = - \frac{3 + 41}{8} :

No Solution

Combine all the solutions

θ = arccos (\frac{- 3 + 41}{8}) + 2 πn, θ = 2 π - arccos (\frac{- 3 + 41}{8}) + 2 πn

Show solutions in decimal form

θ = 1.13140 \dots + 2 πn, θ = 2 π - 1.13140 \dots + 2 πn

cos (x) = sin (2 x) + cos (3 x)

tan (x) sec (x) - 2 tan (x) = 0

- 9 cos^{2} (θ) + 9 = 17 sin (θ) - 8

θ, r = 2 cos (θ)

sin (2 x) = \frac{2 2}{3}

What is the general solution for (18cos(2θ)+27cos(θ))/2 =0 ?
The general solution for (18cos(2θ)+27cos(θ))/2 =0 is θ=1.13140…+2pin,θ=2pi-1.13140…+2pin