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I'm trying to resolve a trigonometrical exercise. I have two ways to resolve it and I receive two different answers. If you could help explain me why one way is a wrong way to resolve it (without doing reference to the trigonometric functions please).

Using double-angle formulas one way is: \begin{align} \cos(2x) = \cos^2x \\ \ 2\cos^2x -1 = \cos^2x \\ \ \cos^2x - 1 = 0 \\ \ \cos^2x = 1 \\ \ x = 360^\circ k \end{align} The second way is: \begin{align} \cos(2x) = \cos^2x \\ \ \cos^2x - \sin^2x = \cos^2x \\ \ -\sin^2x = 0 \\ \ \sin^2x = 0\\ \ x = 180^\circ k \end{align} As you see there are two answers, I know that the last one is the right one but don't know why (please don't do reference to the graphs).

Dominik
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Pichi Wuana
  • 923

3 Answers3

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The first one should have $\cos x=\pm 1\Rightarrow x=n.180$

David Quinn
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In your first calculations, the numbers $x = 360^\circ k$ only yield the solution to the equation $\cos(x) = 1$. However, the equation $\cos^2(x) = 1$ has other solutions, namely the $x$ that satisfy $\cos(x) = -1$. Therefore in the first case you also have $x = 360^\circ k + 180^\circ$ as a solution, therefore all solutions have the form $x = 180^\circ k$.

Dominik
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Technically the second way is correct. But this is only true because you accidentally messed up the solution to the equation $\cos^2({x}) = 1$. In fact, this equation is only solved when $x=180^\circ k$.

Therefore, both $methods$ are correct. But only one solution is correct: $x=180^\circ k$.

NoseKnowsAll
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