Question

Find the general solution of each of the equations:

$\left(i\right)sin2x=-\frac{1}{2}$

(ii)tan 3x =-1

Answer 1

$\left(i\right)sin2x=-\frac{1}{2}=-sin\frac{\pi }{6}=sin(\pi +\frac{\pi }{6})=sin\frac{7\pi }{6}$

$\Rightarrow sin2x=sin\frac{7\pi }{6}$

$\Rightarrow 2x=\{n\pi +(-1{)}^n.\frac{7\pi }{6}\}$, where $n\in I$

$\Rightarrow x=\{\frac{n\pi }{2}+(-1{)}^n.\frac{7\pi }{12}\}$, where $n\in I.$

Hence, the general solution is $x=\{\frac{n\pi }{2}+(-1{)}^n.\frac{7\pi }{12}\}$, where $n\in I.$

$\left(ii\right)tan3x=-1=-tan\frac{\pi }{4}=tan(\pi -\frac{\pi }{4})=tan\frac{3\pi }{4}$

$\Rightarrow tan3x=tan\frac{3\pi }{4}$

$\Rightarrow sin4x=0$ or $cos2x=cos\frac{2\pi }{3}$

$\Rightarrow 4x=n\pi$ or $2x=(2m\pi \pm \frac{2\pi }{3})$, where m, $n\in I.$

$\Rightarrow x=\frac{n\pi }{4}$ or $x=(m\pi \pm \frac{\pi }{3})m$ where m, $n\in I.$

Hence, the general solution is given by $x=\frac{n\pi }{4}$ or $x=(m\pi \pm \frac{\pi }{3})$, where m, $n\in I$.