Question

sinx>cosx

решите неравенство

Answer 1

$\displaystyle\bf\\Sinx > Cosx\\\\Sinx-Cosx > 0\\\\Sinx-Sin\Big(\frac{\pi }{2} -x\Big) > 0\\\\\\2Sin\frac{x-\frac{\pi }{2}+x }{2} \cdot Cos\frac{x+\frac{\pi }{2} -x}{2} > 0\\\\\\2Sin\Big(x-\frac{\pi }{4} \Big) Cos\frac{\pi }{4} > 0\\\\\\2Sin\Big(x-\frac{\pi }{4} \Big) \cdot\frac{\sqrt{2} }{2} > 0\\\\\\\sqrt{2} Sin\Big(x-\frac{\pi }{4} \Big) > 0\\\\\\Sin\Big(x-\frac{\pi }{4} \Big) > 0\\\\\\2\pi n < x-\frac{\pi }{4} < \pi +2\pi n ,n\in Z$

$\displaystyle\bf\\\boxed{\frac{\pi }{4} +2\pi n < x < \frac{5\pi }{4} +2\pi n,n\in Z}$

Answer 2

Ответ:

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Объяснение:

sinx - cosx > 0 | · $\frac{\sqrt{2} }{2}$  ⇔ sinx · $\frac{\sqrt{2} }{2}$ - cosx · $\frac{\sqrt{2} }{2}$ >0 ⇔ sinx · cos$\frac{\pi }{4}$ - cosx · sin$\frac{\pi }{4}$ >0

⇔ sin(x - $\frac{\pi }{4}$) >0  ⇔ 2πk + π > x - $\frac{\pi }{4}$ > 0 + 2πk ⇔ 2πk + $\frac{5\pi }{4}$ > x > $\frac{\pi }{4}$ + 2πk ; k ∈ Z