# What are the values of $x$ such that $\cos x \lessgtr \sin x$?

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?

• Hint: You can get this information from the unit circle directly. Commented May 3, 2021 at 7:35
• 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. Commented May 3, 2021 at 8:18
• 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.
– zwim
Commented May 3, 2021 at 12:42

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.