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Question Number 16092 by Tinkutara last updated on 17/Jun/17

Find the set of values of x \in [0, 2 π]

which satisfy \sin x > \cos x .

(1) (\frac{π}{4}, \frac{3 π}{4}) \cup (\frac{5 π}{4}, 2 π)

(2) (0, \frac{π}{4}) \cup (\frac{5 π}{4}, 2 π)

(3) (\frac{π}{4}, \frac{5 π}{4})

(4) (0, \frac{3 π}{4}) \cup (\frac{5 π}{4}, 2 π)

Commented by Tinkutara last updated on 19/Jun/17

Thanks Sir!

Commented by mrW1 last updated on 17/Jun/17

(3)

Commented by Tinkutara last updated on 18/Jun/17

Can you explain please ?

Commented by mrW1 last updated on 18/Jun/17

\sin x - \cos x > 0

\sqrt{2} (\frac{1}{\sqrt{2}} \sin x - \frac{1}{\sqrt{2}} \cos x) > 0

\sqrt{2} \sin (x - \frac{π}{4}) > 0

\sin (x - \frac{π}{4}) > 0

\Rightarrow 0 < x - \frac{π}{4} < π

\frac{π}{4} < x < \frac{5 π}{4}

Commented by mrW1 last updated on 18/Jun/17

or you just have a look at the graphs

from \sin x and \cos x .

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