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How can I find all values of $x \in [0, 2\pi]$ such that $\cos x > \sin x$ and, similarly, considering the same interval, the values which satisfy $\cos x < \sin x$?

My initial attempt was to divide both sides of the inequality by $\cos x$: $$\tan x < 1$$

What should I do next? Is this path correct?

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    $\begingroup$ Hint: You can get this information from the unit circle directly. $\endgroup$ Commented May 3, 2021 at 7:35
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    $\begingroup$ You can divide both sides of an inequality only if you know the divisor is positive. If the divisor is negative, you have to reverse the inequality. $\endgroup$
    – Bernard
    Commented May 3, 2021 at 8:18
  • $\begingroup$ Instead of dividing that causes inequalities to reverse when $\cos x$ changes sign, better use $\cos(x)-\sin(x)=\sqrt{2}\sin(x+\phi)$ and it is quite easy to study $\sin(x+\phi)$ sign. $\endgroup$
    – zwim
    Commented May 3, 2021 at 12:42

1 Answer 1

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HINT: Be careful with the division on both sides, as $\cos(x)$ changes sign in the interval you are considering. Actually, even $\sin(x)$ changes sign.

You could also try to solve the inequality with a graphical method, plotting the two functions $\sin(x)$ and $\cos(x)$ on a paper and seeing where one of them is greater than the other.

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