Key to "Find the solution of cos⁡(2x)−cos⁡(x)=0 in the interval..."

Talamancoeb

Answered question

2021-12-17

Find the solution of

\cos (2 x) - \cos (x) = 0

in the interval

[0; 2 π)

nghodlokl

Beginner2021-12-18Added 33 answers

We remember that

\cos (2 x) = \cos^{2} (x) - \sin^{2} (x)

And we have

\cos^{2} (x) - \sin^{2} (x) - \cos (x) = 0

Now, get everything in terms of cosine:

\sin^{2} (x) + \cos^{2} (x) = 1

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

Therefore,

\cos^{2} (x) - (1 - \cos^{2} (x)) - \cos (x) = 0

2 \cos^{2} (x) - \cos (x) - 1 = 0

(\cos (x) - 1) (2 \cos (x) + 1) = 0

Now, solve

\cos (x) - 1 = 0

2 \cos (x) + 1 = 0

\cos (x) = 1

Implies that

x = 0

, as we're working in

[0; 2 π)

. And it's the only value yielding one for cosine.

2 \cos (x) = - 1

\cos (x) = - \frac{1}{2}

Implies that

x = \frac{2 π}{3}, \frac{4 π}{3}

as these values return

- \frac{1}{2}

for cosine in the given interval.

Ethan Sanders

Beginner2021-12-19Added 35 answers

\cos 2 x - \cos x = 0 \Leftrightarrow 2 \cos^{2} x - \cos x - 1 = 0 \Leftrightarrow (2 \cos x + 1) (\cos x - 1) = 0 \Rightarrow \cos x = - \frac{1}{2} and \cos x = 1

\cos x = \frac{1}{2} \Leftrightarrow x = \frac{2 π}{3}; x = \frac{4 π}{3}

\cos x = 1 \Leftrightarrow x = 0

2 π

is not included in given interval

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