Question

4sin²x + 2sinx (_/3 -1) - _/3. Find its general solution​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\sf{4sin^{2}(x)+2(\sqrt{3}-1)sin(x)}-\sqrt{3}=0$

$\sf{\implies\,sin(x)=\dfrac{-2(\sqrt{3}-1)\pm\sqrt{\{2(\sqrt{3}-1)\}^{2}+4\cdot4\cdot\sqrt{3}} }{2\cdot4}$

$\sf{\implies\,sin(x)=\dfrac{-2(\sqrt{3}-1)\pm\sqrt{4(\sqrt{3}-1)^{2}+16\sqrt{3}} }{8}$

$\sf{\implies\,sin(x)=\dfrac{-2(\sqrt{3}-1)\pm2\sqrt{(\sqrt{3}-1)^{2}+4\sqrt{3}} }{8}$

$\sf{\implies\,sin(x)=\dfrac{-2(\sqrt{3}-1)\pm2\sqrt{(\sqrt{3}+1)^{2}} }{8}$

$\sf{\implies\,sin(x)=\dfrac{-2(\sqrt{3}-1)\pm2(\sqrt{3}+1) }{8}$

$\sf{\implies\,sin(x)=\dfrac{-2(\sqrt{3}-1)+2(\sqrt{3}+1) }{8}\,\,\,\,or\,\,\,\, sin(x)=\dfrac{-2(\sqrt{3}-1)-2(\sqrt{3}+1) }{8}}$

$\sf{\implies\,sin(x)=\dfrac{-2\sqrt{3}+2+2\sqrt{3}+2 }{8}\,\,\,\,or\,\,\,\, sin(x)=\dfrac{-2\sqrt{3}+2-2\sqrt{3}-2 }{8}}$

$\sf{\implies\,sin(x)=\dfrac{2+2 }{8}\,\,\,\,or\,\,\,\, sin(x)=\dfrac{-2\sqrt{3}-2\sqrt{3} }{8}}$

$\sf{\implies\,sin(x)=\dfrac{4}{8}\,\,\,\,or\,\,\,\, sin(x)=\dfrac{-4\sqrt{3} }{8}}$

$\sf{\implies\,sin(x)=\dfrac{1}{2}\,\,\,\,or\,\,\,\, sin(x)=-\dfrac{\sqrt{3} }{2}}$

$\sf{\implies\,sin(x)=sin\bigg(\dfrac{\pi}{6}\bigg)\,\,\,\,or\,\,\,\, sin(x)=sin\bigg(-\dfrac{\pi }{3}}\bigg)$

$\sf{\implies\,x=n\pi+(-1)^{n}\dfrac{\pi}{6}\,\,\,\,or\,\,\,\,x=m\pi+(-1)^{m}\bigg(-\dfrac{\pi}{3}\bigg)}$

$\sf{\implies\,x=n\pi+(-1)^{n}\cdot\dfrac{\pi}{6}\,\,\,\,or\,\,\,\,x=m\pi+(-1)^{m+1}\cdot\dfrac{\pi}{3}}$