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bija089 [108]
2 years ago
9

Find all solutions to the equation in the interval [0,2pi). Enter the solutions in increasing order. Cos 2x = cos x

Mathematics
1 answer:
goblinko [34]2 years ago
4 0

Step-by-step explanation:

\cos(2x)  =  \cos(x)

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

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

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

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

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

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

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

2 \cos(x)  + 1 = 0

2 \cos(x)  =  - 1

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

x =   \frac{2\pi}{3} ,x =  \frac{4\pi}{3}

\cos(x)   - 1 = 0

\cos(x)  = 1

x = 0

So our answer are

0, \frac{2\pi}{3} , \frac{4\pi}{3}

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