Expert Answer to "find the x: (cos⁡3x+cos⁡4x)(cos⁡3x+cos⁡x)=14"

Aliyah Mendez

Answered question

2022-04-07

find the x:

(\cos 3 x + \cos 4 x) (\cos 3 x + \cos x) = \frac{1}{4}

gil001q4wq

Beginner2022-04-08Added 11 answers

Let

\cos x = t

.
Thus, we get

(4 t^{3} - 3 t + 8 t^{4} - 8 t^{2} + 1) (4 t^{3} - 3 t + t) = \frac{1}{4}

(2 t + 1) (4 t^{2} + 2 t - 1) (16 t^{4} - 8 t^{3} - 16 t^{2} + 8 t + 1) = 0

16 t^{4} - 8 t^{3} - 16 t^{2} + 8 t + 1 = 0

we can solve by the following way

{(4 t^{2} - t - \frac{3}{2})}^{2} - 5 {(t - \frac{1}{2})}^{2} = 0 .

I think the rest is smooth.
I think the following way a bit of better.
After using of your work we need to solve

16 \sin \frac{x}{2} \cos \frac{x}{2} \cos {x} \cos 2 x \cos \frac{7 x}{2} = \sin \frac{x}{2}

2 \sin 4 x \cos \frac{7 x}{2} = \sin \frac{x}{2}

\sin \frac{15 x}{2} + \sin \frac{x}{2} = \sin \frac{x}{2}

\sin \frac{15 x}{2} = 0

Now, we need to delete all roots of

\sin \frac{x}{2} = 0

and to write the answer.

wyjadaczeqa8

Beginner2022-04-09Added 14 answers

Substituting

\cos (t) = \frac{e^{i t} + e^{- i t}}{2}

and letting

y = e^{i x}

, you get:

(\cos (4 x) + \cos (3 x)) (\cos (3 x) + \cos (x)) = \frac{1}{4}

(y^{4} + y^{- 4} + y^{3} + y^{- 3}) (y^{3} + y^{- 3} + y^{1} + y^{- 1}) = 1

(\sum_{k = - 7}^{7} y^{k}) + 1 = 1

\sum_{k = 0}^{14} y^{k} = 0

\frac{y^{15} - 1}{y - 1} = 0

e^{15 i x} = 1

and

e^{i x} \neq 1

, leaving the solutions being

x = \frac{2 π n}{15}

for

n \in Z

and 15∤n. I suppose it is just the coincidence of contrived problems that it worked out to be a geometric series.

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