Question

Solve the following equations:
(i) ${\sin }^2\mathrm{\theta }-\cos \mathrm{\theta }=\frac{1}{4}$
(ii) $2{\cos }^2\mathrm{\theta }-5\cos \mathrm{\theta }+2=0$
(iii) $2{\sin }^{2}x+\sqrt{3}\cos x+1=0$
(iv) $4{\sin }^2\mathrm{\theta }-8\cos \mathrm{\theta }+1=0$
(v) ${\tan }^{2}x+\left(1-\sqrt{3}\right)\tan x-\sqrt{3}=0$
(vi) $3{\cos }^2\mathrm{\theta }-2\sqrt{3}\sin \mathrm{\theta }\cos \mathrm{\theta }-3{\sin }^2\mathrm{\theta }=0$
(vii) $\cos 4\mathrm{\theta }=\cos 2\mathrm{\theta }$

Answer 1

(i) ${\sin }^2\theta -\cos \theta =\frac{1}{4}$
$$\begin{gathered} \Rightarrow 1-{\cos }^2\theta -\cos \theta =\frac{1}{4} \\ \Rightarrow 4-4{\cos }^2\theta -4\cos \theta =1 \\ \Rightarrow 4{\cos }^2\theta +4\cos \theta -3=0 \\ \Rightarrow 4{\cos }^2\theta +6\cos \theta -2\cos \theta -3=0 \\ \Rightarrow 2\cos \theta (2\cos \theta +3)-1(2\cos \theta +3)=0 \\ \Rightarrow (2\cos \theta +3)(2\cos \theta -1)=0 \end{gathered}$$

$\Rightarrow (2\cos \theta -1)=0$ or $2\cos \theta +3=0$
$\Rightarrow \cos \theta =\frac{1}{2}$ or $\cos \theta =-\frac{3}{2}$

$\cos \theta =-\frac{3}{2}$ is not possible.
$$\begin{gathered} \therefore \cos \theta =\frac{1}{2} \\ \Rightarrow \cos \theta =\cos \frac{\mathrm{\pi }}{3} \\ \Rightarrow \theta =2n\pi \pm \frac{\mathrm{\pi }}{3},n\in Z \end{gathered}$$

(ii)
$$\begin{gathered} 2{\cos }^2\theta -5\cos \theta +2=0 \\ \Rightarrow 2{\cos }^2\theta -4\cos \theta -\cos \theta +2=0 \\ \Rightarrow 2\cos \theta (\cos \theta -2)-1(\cos \theta -2)=0 \\ \Rightarrow (\cos \theta -2)(2\cos \theta -1)=0 \end{gathered}$$

$\Rightarrow (\cos \theta -2)=0$ or, $(2\cos \theta -1)=0$
$\cos \theta =2$ is not possible.

$$\begin{gathered} \therefore 2\cos \theta -1=0 \\ \Rightarrow \cos \theta =\frac{1}{2} \\ \Rightarrow \cos \theta =\cos \frac{\mathrm{\pi }}{3} \\ \Rightarrow \theta =2n\pi \pm \frac{\mathrm{\pi }}{3},n\in Z \end{gathered}$$

(iii)
$$\begin{gathered} 2{\sin }^{2}x+\sqrt{3}\cos x+1=0 \\ \Rightarrow 2-2{\cos }^{2}x+\sqrt{3}\cos x+1=0 \\ \Rightarrow 2{\cos }^{2}x-\sqrt{3}\cos x-3=0 \\ \Rightarrow 2{\cos }^{2}x-2\sqrt{3}\cos x+\sqrt{3}\cos x-3=0 \\ \Rightarrow 2\cos x(\cos x-\sqrt{3})+\sqrt{3}(\cos x-\sqrt{3})=0 \\ \Rightarrow (2\cos x+\sqrt{3})(\cos x-\sqrt{3})=0 \end{gathered}$$

$\Rightarrow$ $(2\cos x+\sqrt{3})=0$ or $(\cos x-\sqrt{3})=0$

$\cos x=\sqrt{3}$ is not possible.
$$\begin{gathered} \therefore 2\cos x+\sqrt{3}=0 \\ \Rightarrow \cos x=-\frac{\sqrt{3}}{2} \\ \Rightarrow \cos x=\cos \frac{5\mathrm{\pi }}{6} \\ \Rightarrow x=2n\mathrm{\pi }\pm \frac{5\mathrm{\pi }}{6},\mathrm{n}\in \end{gathered}$$


(iv)
$4{\sin }^2\theta -8\cos \theta +1=0$
$$\begin{gathered} \Rightarrow 4-4{\cos }^2\theta -8\cos \theta +1=0 \\ \Rightarrow 4{\cos }^2\theta +8\cos \theta -5=0 \\ \Rightarrow 4{\cos }^2\theta +10\cos \theta -2\cos \theta -5=0 \\ \Rightarrow 2\cos \theta (2\cos \theta +5)-1(2\cos \theta +5)=0 \\ \Rightarrow (2\cos \theta -1)(2\cos \theta +5)=0 \end{gathered}$$

$\Rightarrow (2\cos \theta -1)=0$ or $(2\cos \theta +5)=0$
Now,
$2\cos \theta +5=0\Rightarrow \cos \theta =-\frac{5}{2}$ (It is not possible.)
$$\begin{gathered} \therefore 2\cos \theta -1=0 \\ \Rightarrow \cos \theta =\frac{1}{2} \\ \Rightarrow \cos \theta =\cos \frac{\mathrm{\pi }}{3} \\ \Rightarrow \theta =2n\pi \pm \frac{\mathrm{\pi }}{3},n\in Z \end{gathered}$$


(v)
${\tan }^{2}x+(1-\sqrt{3})\tan x-\sqrt{3}=0$
$$\begin{gathered} \Rightarrow {\tan }^{2}x+\tan x-\sqrt{3}\tan x-\sqrt{3}=0 \\ \Rightarrow \tan x(\tan x+1)-\sqrt{3}(\tan x+1)=0 \\ \Rightarrow (\tan x-\sqrt{3})(\tan x+1)=0 \end{gathered}$$

$\Rightarrow (\tan x-\sqrt{3})=0$ or $(\tan x+1)=0$
Now,

$$\begin{gathered} \tan x-\sqrt{3}=0 \\ \Rightarrow \tan x=\sqrt{3} \\ \Rightarrow \tan x=\tan \frac{\mathrm{\pi }}{3} \\ \Rightarrow x=n\mathrm{\pi }+\frac{\mathrm{\pi }}{3},n\in Z \end{gathered}$$
And,

$$\begin{gathered} \tan x=-1 \\ \Rightarrow \tan x=\tan \left(-\frac{\mathrm{\pi }}{4}\right) \\ \Rightarrow x=m\mathrm{\pi }-\frac{\mathrm{\pi }}{4},\mathrm{m}\in \mathrm{Z} \end{gathered}$$


(vi)
$3{\cos }^2\theta -2\sqrt{3}\sin \theta \cos \theta -3{\sin }^2\theta =0$
Now,
$$\begin{gathered} 3({\cos }^2\theta -{\sin }^2\theta )-\sqrt{3}\sin 2\theta =0 \\ \Rightarrow 3\cos 2\theta -\sqrt{3}\sin 2\theta =0 \\ \Rightarrow \sqrt{3}(\sqrt{3}\cos 2\theta -\sin 2\theta )=0 \\ \Rightarrow (\sqrt{3}\cos 2\theta -\sin 2\theta )=0 \\ \Rightarrow \frac{\sin 2\theta }{\cos 2\theta }=\sqrt{3} \\ \Rightarrow \tan 2\theta =\tan \frac{\mathrm{\pi }}{3} \\ \Rightarrow 2\theta =n\pi +\frac{\mathrm{\pi }}{3},n\in Z \\ \Rightarrow \theta =\frac{n\mathrm{\pi }}{2}+\frac{\mathrm{\pi }}{6},n\in Z \end{gathered}$$



(vii)

$$\begin{gathered} \cos 4\theta =\cos 2\theta \\ \Rightarrow 4\theta =2n\pi \pm 2\theta ,n\in Z \end{gathered}$$

On taking positive sign, we have:

$$\begin{gathered} 4\theta =2n\pi +2\theta \\ \Rightarrow 2\theta =2n\pi \\ \Rightarrow \theta =n\pi ,n\in Z \end{gathered}$$
On taking negative sign, we have:

$$\begin{gathered} 4\theta =2n\pi -2\theta \\ \Rightarrow 6\theta =2n\pi \\ \Rightarrow \theta =\frac{n\mathrm{\pi }}{3},n\in Z \end{gathered}$$