Question

When $0^{\circ }<\theta <{90}^{\circ }$, Solve the following equations :

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

(ii) $3\cos \theta =2{\sin }^2\theta$

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

(iv)${\tan }^2\theta =3(\sec \theta -1)$.

Answer 1

A) $2{\cos }^2\theta +\sin \theta -2=0$

or, $2-2{\sin }^2\theta +\sin \theta -2-0$

or, $\sin \theta (-2\sin \theta +1)=0$

Since $\theta \ne 0$, $\sin \theta =\frac{1}{2}\Rightarrow \theta ={30}^o$. [Since $0^o<\theta <{90}^o$]

B) $3\cos \theta =2{\sin }^2\theta$

or, $3\cos \theta =2-2{\cos }^2\theta$

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

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

Since $\cos \theta \ne -2$, $\cos \theta =\frac{1}{2}\Rightarrow \theta ={60}^o$ [Since $0^o<\theta <{90}^o$]

C) ${\sec }^2\theta -2\tan \theta =0$

or, $1+{\tan }^2\theta -2\tan \theta =0$

or, $(\tan \theta -1{)}^2=0$

$\Rightarrow \tan \theta =1\Rightarrow \theta ={45}^o$[Since $0^o<\theta <{90}^o$]

D) ${\tan }^2\theta =3(\sec \theta -1)$

or, ${\sec }^2\theta -1=3(\sec \theta -1)$

or, ${\sec }^2\theta -3\sec \theta +2=0$

or, $(\sec \theta -1)(\sec \theta -2)=0$

Since $0^o<\theta <{90}^o$, $\sec \theta =2\Rightarrow \theta ={60}^0$.