olve in real numbers the equation
\displaystyle \-x+sqrt{1-x^{2}}\-=sqrt{2}\left(2x^{2}-1\right)\-x+sqrt1−x
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Answer:
Substitutex=cos\alp,\alp∈[0:;π]
\displaystyle Then\; |x+\sqrt{1-x^2}|=\sqrt{2}(2x^2-1)\Leftright |cos\alp +sin\alp |=\sqrt{2}(2cos^2\alp -1)Then∣x+1−x2∣=2(2x2−1)\Leftright∣cos\alp+sin\alp∣=2(2cos2\alp−1)
\displaystyle |\N {\sqrt{2}}cos(\alp-\frac{\pi}{4})|=\N {\sqrt{2}}cos(2\alp )\Right \alp\in[0\: ;\: \frac{\pi}{4}]\cup [\frac{3\pi}{4}\: ;\: \pi]∣N2cos(\alp−4π)∣=N2cos(2\alp)\Right\alp∈[0;4π]∪[43π;π]
1) \displaystyle \alp \in [0\: ;\: \frac{\pi}{4}]\alp∈[0;4π]
\displaystyle cos(\alp -\frac{\pi}{4})=cos(2\alp )\dotscos(\alp−4π)=cos(2\alp)…
2. \displaystyle \alp\in [\frac{3\pi}{4}\: ;\: \pi]\alp∈[43π;π]
\displaystyle -cos(\alp -\frac{\pi}{4})=cos(2\alp )\dots−cos(\alp−4π)=cos(2\alp)…
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Answer:
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