The answers from the back of the book says the answers are $\pi/2$ and $3\pi/2$. I thought that you can use the double angle formula in this problem, but it doesn't seem right?
$\endgroup$ 43 Answers
$\begingroup$As you suggest, since $\cos 2x = \cos^2 x - \sin^2 x$, this is equivalent to the equation $\cos 2x = 1$, which has solutions when $2x$ is an integer multiple of $2\pi$, so the solutions are $$x = \dots, -2\pi, -\pi, 0, \pi, 2\pi, \dots$$ So the book has the wrong answer.
$\endgroup$ $\begingroup$$\cos(2x) = \cos^{2}(x) - \sin^{2}(x)$, So $\cos(2x) = 1$ has solutions of the form $2x = 2k\pi$ for $k \in \Bbb{Z}$ That is $x = k\pi$ where $k \in \Bbb{Z}$.
$\endgroup$ $\begingroup$$$\cos^2x-\sin^2x-1=0$$
Recall the idenity
$$\cos^2x+\sin^2x=1$$
$$\sin^2x=1-\cos^2x$$
$$\cos^2x-(1-\cos^2x)=1$$
$$2\cos^2x=2$$
$$\cos^2x=1$$
You may proceed with these,
Also notice that
$$\cos^2x-1=\sin^2x$$
$$\sin^2x=0$$
$\endgroup$
