Question

Find the general solution of the following equations :
(i) $2{\cos }^2\theta -5\cos \theta +2=0$
(ii) $4{\cos }^2\theta -4\sin \theta =1$
(iii) $2{\sin }^2\theta +3\cos \theta =0$
(iv) ${\tan }^2\theta -4\sec \theta +5=0$

Answer 1

i) ${2\cos }^2\theta -5\cos \theta +2=0$

$\Rightarrow 2{\cos }^2\theta +4\cos \theta -\cos \theta +2=0$

$\Rightarrow (2\cos \theta -1)(\cos \theta -2)=0$

$\therefore$ $\cos \theta =1/2⟶\left(1\right)$ or $\cos \theta =2⟶\left(2\right)$

Now if $\cos \theta =\cos \alpha$ $\cos \theta =2$ is not possible

when general solution is $\because$ $cos\theta ϵ[-1,1]$

$\theta =2n\pi \pm \alpha$

$\therefore$ $\cos \theta =1/2=\cos \pi /3$

$\theta =2n\pi \pm \pi /3$

$nϵZ$

ii) $4{\cos }^2\theta -4\sin \theta =1$

$\Rightarrow 4(1-{\sin }^2\theta )-4\sin \theta =1$ $(\because {\sin }^2\theta +{\cos }^2\theta =1)$

$\Rightarrow 4-4{\sin }^2\theta -4\sin \theta =1$

$\Rightarrow 4{\sin }^2\theta +4\sin \theta -3=0$

$\Rightarrow 4{\sin }^2\theta +6\sin \theta -2\sin \theta -3=0$

$\Rightarrow 2\sin \theta (2\sin \theta +3)-1(2\sin \theta +3)=0$

$\Rightarrow (2\sin \theta -1)(2\sin \theta +3)$

$\sin \theta =1/2⟶\left(1\right)$ or $\sin \theta =-3/2⟶\left(2\right)$

if $sinx=siny$ but $\sin \theta =-3/2$ is

the $x=n\pi +{(-1)}^{n}y;nϵZ$ not possible since

$\therefore$ $\sin \theta =1/2=\sin \pi /6$ $\sin \theta ϵ[-1,1]$

$\therefore$ $\theta =n\pi +{(-1)}^n\pi /6$

iii) $2{\sin }^2\theta +3\cos \theta =0$

$\Rightarrow 2{\cos }^2\theta -3\cos \theta -2=0$

$\Rightarrow 2{\cos }^2\theta -4\cos \theta +\cos \theta -2=0$

$\Rightarrow (2\cos \theta +1)(\cos \theta -2)=0$

$\therefore$ $\cos \theta =-1/2$ or $\cos \theta =2$

$\theta =2n\pi \pm (-\pi /3)$ $cos\theta =2$ is not possible

iv) ${\tan }^2\theta -4\sec \theta +5=0$ $\because$ $\cos \theta ϵ[-1,1]$

$\Rightarrow {\sec }^2\theta -4\sec \theta +4=0$

$\Rightarrow {(\sec \theta -2)}^2=0$

$\Rightarrow \sec \theta =2\Rightarrow \cos \theta =1/2$

$\therefore$ general solution : $\theta =2n\pi \pm \pi /3$