Question

Решите уравнение
√2sin(2x+pi/4)–√3sinx=sin2x+1

Answer 1

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Answer 2

$\sqrt2\, sin(2x+\frac{\pi}{4})-\sqrt3\, sinx=sin2x+1\\\\\sqrt2\cdot (sin2x\cdot cos\frac{\pi}{4}+cos2x\cdot sin\frac{\pi}{4})-\sqrt3\, sinx=sin2x+1\\\\\sqrt2\cdot (sin2x\cdot \frac{\sqrt2}{2}+cos2x\cdot \frac{\sqrt2}{2})-\sqrt3\, sinx=sin2x+1\\\\\underline {sin2x}+cos2x-\sqrt3sinx=\underline {sin2x}+1\\\\(\underbrace {1-2sin^2x}_{cos2x})-\sqrt3sinx=1\\\\2sin^2x+\sqrt3sinx=0\\\\sinx\cdot (2sinx+\sqrt3)=0\\\\a)\; \; sinx=0\; \; ,\; \; x=\pi n\; ,\; n\in Z$

$b)\; \; sinx=-\frac{\sqrt3}{2}\; \; ,\; \; x=(-1)^{k}\cdot (-\frac{\pi}{3})+\pi k\; ,\; k\in Z\\\\x=(-1)^{k+1}\cdot \frac{\pi}{3}+\pi k\; ,\; k\in Z\\\\Otvet:\; \; x=\pi n\; ,\; \; x=(-1)^{k+1}\cdot \frac{\pi}{3}+\pi k\; ,\; \; n,k\in Z\; .$

$x\in [-\frac{3\pi}{2}\, ;\, 0\, ]:\; \; x=0\; ,\; -\frac{\pi}{3}\; ,\; -\frac{2\pi }{3}\; ,\; -\pi  \; .$