Find the solution of cos⁡(2x)−cos⁡(x)=0 in the interval [0;2π)

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nghodlokl · Accepted Answer

We remember that cos⁡(2x)=cos2(x)−sin2(x)And we havecos2(x)−sin2(x)−cos⁡(x)=0Now, get everything in terms of cosine:sin2(x)+cos2(x)=1sin2(x)=1−cos2(x)Therefore,cos2(x)−(1−cos2(x))−cos⁡(x)=02cos2(x)−cos⁡(x)−1=0(cos⁡(x)−1)(2cos⁡(x)+1)=0Now, solvecos⁡(x)−1=02cos⁡(x)+1=0cos⁡(x)=1Implies that x=0, as we&#039;re working in [0;2π). And it&#039;s the only value yielding one for cosine.2cos⁡(x)=−1cos⁡(x)=−12Implies that x=2π3,4π3 as these values return −12 for cosine in the given interval.

Ethan Sanders · Accepted Answer

cos⁡2x−cos⁡x=0⇔2cos2x−cos⁡x−1=0⇔(2cos⁡x+1)(cos⁡x−1)=0⇒cos⁡x=−12 and cos⁡x=1    cos⁡x=12⇔x=2π3;x=4π3    cos⁡x=1⇔x=0    2π is not included in given interval

Find the solution of cos⁡(2x)−cos⁡(x)=0 in the interval [0;2π)

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