Giải phương trình a)sin2x + sin6x =3 cos² 2x b)sin²x =cos² 2x + cos² 3x

1 bình luận về “Giải phương trình a)sin2x + sin6x =3 cos² 2x b)sin²x =cos² 2x + cos² 3x”

1. Giải đáp + Lời giải và giải thích chi tiết:

   $\begin{array}{l}a)\sin 2x+\sin 6x=3{\cos }^{2}2x\\ \iff \sin 2x+3\sin 2x–4{\sin }^{3}2x–3{\cos }^{2}2x=0\\ \iff 4\sin 2x\left(1–{\sin }^{2}2x\right)–3{\cos }^{2}2x=0\\ \iff 4\sin 2x{\cos }^{2}2x–3{\cos }^{2}2x=0\\ \iff {\cos }^{2}2x\left(4\sin 2x–3\right)=0\\ \iff [\begin{array}{c}\cos 2x=0\\ \sin 2x={\displaystyle \frac{3}{4}}\end{array}\\ \iff [\begin{array}{c}2x={\displaystyle \frac{\pi }{2}}+k\pi \\ 2x=\arcsin \left({\displaystyle \frac{3}{4}}\right)+k2\pi \\ 2x=\pi –\arcsin \left({\displaystyle \frac{3}{4}}\right)+k2\pi \end{array}\left(k\in Z\right)\\ \iff [\begin{array}{c}x={\displaystyle \frac{\pi }{4}}+k{\displaystyle \frac{\pi }{2}}\\ x={\displaystyle \frac{1}{2}}\arcsin \left({\displaystyle \frac{3}{4}}\right)+k\pi \\ x={\displaystyle \frac{\pi }{2}}–{\displaystyle \frac{1}{2}}\arcsin \left({\displaystyle \frac{3}{4}}\right)+k\pi \end{array}\left(k\in Z\right)\end{array}$

   Vậy phương trình có các họ nghiệm là:

   $x={\displaystyle \frac{\pi }{4}}+k{\displaystyle \frac{\pi }{2}};x={\displaystyle \frac{1}{2}}\arcsin \left({\displaystyle \frac{3}{4}}\right)+k\pi ;x={\displaystyle \frac{\pi }{2}}–{\displaystyle \frac{1}{2}}\arcsin \left({\displaystyle \frac{3}{4}}\right)+k\pi \left(k\in Z\right)$

   $\begin{array}{l}b){\sin }^{2}x={\cos }^{2}2x+{\cos }^{2}3x\\ \iff {\left(2{\cos }^{2}x–1\right)}^2+{\left(4{\cos }^{3}x–3\cos x\right)}^2–{\sin }^{2}x=0\\ \iff {\left(2{\cos }^{2}x–1\right)}^2+{\left(4{\cos }^{3}x–3\cos x\right)}^2+{\cos }^{2}x–1=0\\ \iff 16{\cos }^{6}x–20{\cos }^{4}x+6{\cos }^{2}x=0\\ \iff 2{\cos }^{2}x\left(8{\cos }^{4}x–10{\cos }^{2}x+3\right)=0\\ \iff 2{\cos }^{2}x\left(2{\cos }^{2}x–1\right)\left(4{\cos }^2–3\right)=0\\ \iff [\begin{array}{c}\cos x=0\\ \cos x={\displaystyle \frac{1}{\sqrt{2}}}\\ \cos x={\displaystyle \frac{–1}{\sqrt{2}}}\\ \cos x={\displaystyle \frac{\sqrt{3}}{2}}\\ \cos x={\displaystyle \frac{–\sqrt{3}}{2}}\end{array}\iff [\begin{array}{c}x={\displaystyle \frac{\pi }{2}}+k\pi \\ x={\displaystyle \frac{\pi }{4}}+k{\displaystyle \frac{\pi }{2}}\\ x={\displaystyle \frac{\pi }{6}}+k2\pi \\ x={\displaystyle \frac{–\pi }{6}}+k2\pi \\ x={\displaystyle \frac{5\pi }{6}}+k2\pi \\ x={\displaystyle \frac{–5\pi }{6}}+k2\pi \end{array}\left(k\in Z\right)\end{array}$

   Vậy các họ nghiệm của phương trình là: $x={\displaystyle \frac{\pi }{2}}+k\pi ;x={\displaystyle \frac{\pi }{4}}+k{\displaystyle \frac{\pi }{2}};x={\displaystyle \frac{\pi }{6}}+k2\pi ;x={\displaystyle \frac{–\pi }{6}}+k2\pi ;x={\displaystyle \frac{5\pi }{6}}+k2\pi ;x={\displaystyle \frac{–5\pi }{6}}+k2\pi \left(k\in Z\right)$

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