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√2tanxcosx−tanx=0 find all solutions (0,2pie)
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Didn't I just do this one?!
\[tanxsin^2x=tanx\]
plz give me answer i just want to verify
Okay, we've got to the point \(\tan{x}=0 \text{ or }\cos{x}=\frac{1}{\sqrt{2}}\).
ya i got but the radians
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0,2pie
\(\tan{x}=0 \text{ at } x=0 \text{ and } x=\pi\).
inbetween the circle
\(\cos{x}=\frac{1}{\sqrt{2}} \text{ at } x=\frac{\pi}{4} \text{ or } \frac{3\pi}{4}.\)
This is the solution of your first problem.
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coud u explain me second problem pz
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