Question

Find the general solutions of the following equations :
$\left(i\right)sin\theta =\frac{1}{2}\left(ii\right)cos\theta =\frac{-\sqrt{3}}{1}\left(iii\right)cosec\theta =-\sqrt{2}\left(iv\right)sec\theta =\sqrt{2}\left(v\right)tan\theta =-\frac{1}{\sqrt{3}}\left(vi\right)\sqrt{3}sec\theta =2$

Answer 1

We have,
$\left(i\right)sin\theta =\frac{1}{2}\Rightarrow sin\theta =\frac{1}{2}[\because sin\frac{\pi }{6}=\frac{1}{2}]\Rightarrow \text{the general solution is}\theta =n\pi +(-1{)}^n\frac{\pi }{6};n\in z[\because ifsin\theta =sin\alpha \Rightarrow \theta =n\pi +(-1{)}^n\alpha ](ii)cos\theta =\frac{-\sqrt{3}}{1}\text{We have,}\Rightarrow cos\theta =cos(\pi +\frac{\pi }{6})\Rightarrow cos\theta =cos\frac{7\pi }{6}[\because cos\frac{7\pi }{6}=-\frac{\sqrt{3}}{2}]\because \text{the general solution is}\theta =2n\pi =\pm \frac{7\pi }{6},n\in z(iii)cosec\theta =-\sqrt{2}cosec\theta =-\sqrt{2}\Rightarrow \frac{1}{sin\theta }=-\sqrt{2}\Rightarrow sin\theta =-\frac{1}{\sqrt{2}}\Rightarrow sing\theta =sin(\pi +\frac{\pi }{4})\Rightarrow sin\theta =sin\frac{5\pi }{4}orsin\theta =sin(-\frac{\pi }{4})\because sin(-theta)=-sin]theta.\therefore \theta =n\pi +(-1{)}^{n+1}\frac{\pi }{4}.n\in z(iv)sec\theta =\sqrt{2}\text{We have,}sec\theta =\sqrt{2}\Rightarrow \frac{1}{cos\theta }=\sqrt{2}\Rightarrow cos\theta =\frac{1}{\sqrt{2}}\Rightarrow cos\theta =cos\frac{\pi }{4}\Rightarrow \theta =2n\pi \pm \frac{\pi }{4},n\in z(v)tan\theta =-\frac{1}{\sqrt{3}}\text{We have,}tan\theta =-\frac{1}{\sqrt{3}}\Rightarrow tan\theta =-tan\left(\frac{\pi }{6}\right)\Rightarrow tan\theta =tan(-\frac{\pi }{6})[\because tan(-\theta )=-tan\theta ]\Rightarrow \theta =n\pi +(-\frac{\pi }{6}),n\in zor\theta =n\pi -\frac{\pi }{6},n\in z(vi)\sqrt{3}sec\theta =2\text{We have,}\sqrt{3}sec\theta =2\Rightarrow \frac{1}{cos\theta }=\frac{2}{\sqrt{3}}\Rightarrow cos\theta =\frac{\sqrt{3}}{2}\Rightarrow cos\theta =cos\left(\frac{\pi }{6}\right)\Rightarrow \theta =2n\pi \pm \frac{\pi }{6},n\in z$