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Question Number 201728 by hardmath last updated on 11/Dec/23

$${\cos }^{2}4\mathrm{x}\cdot {\sin }^{2}4\mathrm{x}=0,25\mathrm{for}\mathrm{equation}$$$$[0;90]\mathrm{how}\mathrm{many}\mathrm{roots}\mathrm{are}\mathrm{there}\mathrm{in}\mathrm{the}$$$$\mathrm{piece}?$$

Answered by esmaeil last updated on 11/Dec/23

$${(2cos4x.sin4x)}^2=1\to$$$${sin}^{2}8x=1\to \frac{1-cos16x}{2}=1\to$$$$cos16x=-1=cos\pi \to$$$$x=\frac{2k\pi }{16}\pm \frac{\pi }{16}\to x=(\frac{\pi }{16},\frac{3\pi }{16},\frac{5\pi }{16},\frac{7\pi }{16})$$$$$$$$$$$$$$

Commented by hardmath last updated on 11/Dec/23

$$\mathrm{thank}\mathrm{you}\mathrm{dear}\mathrm{professor},$$$$\mathrm{i}.\mathrm{e}.\mathrm{the}\mathrm{piece}[0;90]\mathrm{has}\mathrm{four}\mathrm{roots}?$$$$\mathrm{answer}:4?$$

Commented by esmaeil last updated on 11/Dec/23

$$yes$$

Commented by esmaeil last updated on 11/Dec/23

$$\frac{2k-1}{16}⩽\frac{1}{2}\Rightarrow k⩽4$$

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