Question

Resuelve las siguientes ecuaciones para 0 < x < 2Pi
3. a) 2cos x + 3 = 2
4. b) sen3x - 2 = -3sen3x
5. c) senx(2 - senx) = cos2x

Resuelve las siguientes ecuaciones para 0 < x < 2Pi
6. a) 2cos x + 3 = 2
7. b) sen3x - 2 = -3sen3x
8. c) senx(2 - senx) = cos2x

Answer 1

Hola.

$1) \\ 2\cos \left(x\right)+3=2,\:0\ \textless \ x\ \textless \ 2\pi \\ \\ 2\cos \left(x\right)=2-3 \\ \\ 2\cos \left(x\right)=-1 \\ \\ \cos \left(x\right)=-\frac{1}{2} \\ \\ x=\frac{2\pi }{3}+2\pi n,\:x=\frac{4\pi }{3}+2\pi n \\ \\ \boxed{x=\frac{2\pi }{3},\:x=\frac{4\pi }{3}}$

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$2) \\ \\ \sin \left(3x\right)-2=-3\sin \left(3x\right),\:0\ \textless \ x\ \textless \ 2\pi \\ \\ Sustitucion: \sin \left(3x\right)=u \\ \\ u-2=-3u \\ \\ u=-3u+2 \\ \\ u+3u=2 \\ \\ 4u=2 \\ \\ \frac{4u}{4}=\frac{2}{4} \\ \\ u=\frac{1}{2} \\ \\ Volvemos \ a \ la \ original: \\ \\ \sin \left(3x\right)=\frac{1}{2} \\ \\ \sin \left(3x\right)=\frac{1}{2},\:0\ \textless \ x\ \textless \ 2\pi \\ \\ \boxed{x=\frac{\pi }{18},\:x=\frac{5\pi }{18},\:x=\frac{17\pi }{18},\:x=\frac{29\pi }{18},\:x=\frac{13\pi }{18},\:x=\frac{25\pi }{18}}$

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$3) \\ \\ \sin \left(x\right)\left(2-\sin \left(x\right)\right)=\cos \left(2x\right),\:0\ \textless \ x\ \textless \ 2\pi \\ \\ \sin \left(x\right)\left(2-\sin \left(x\right)\right)-\cos \left(2x\right)=0 \\ \\ Usamos: \cos \left(2x\right)=1-2\sin ^2\left(x\right) \\ \\ -\left(1-2\sin ^2\left(x\right)\right)+\left(2-\sin \left(x\right)\right)\sin \left(x\right)=0 \\ \\ -1+\sin ^2\left(x\right)+2\sin \left(x\right)=0 \\ \\ Sustituimos: \sin \left(x\right)=u \\ \\ -1+u^2+2u=0$

$Ecuacion \ general: \\ \\ \frac{-2+\sqrt{2^2-4\cdot \:1\cdot \left(-1\right)}}{2\cdot \:1} \\ \\ Soluciones: \\ \\ u=\sqrt{2}-1,\:u=-1-\sqrt{2} \\ \\ Cambiamos \ a \ la \ original: \\ \\ \sin \left(x\right)=\sqrt{2}-1,\:\sin \left(x\right)=-1-\sqrt{2} \\ \\ \boxed{x=\pi -\arcsin \left(\sqrt{2}-1\right),\:x=\arcsin \left(\sqrt{2}-1\right)}$

¡Espero haberte ayudado, saludos... G.G.H!