Question

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

Answer 1

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

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

(ii)
$$\begin{gathered} 2{\cos }^{2}x-5\cos x+2=0 \\ \Rightarrow 2{\cos }^{2}x-4\cos x-\cos x+2=0 \\ \Rightarrow 2\cos x(\cos x-2)-1(\cos x-2)=0 \\ \Rightarrow (\cos x-2)(2\cos \theta -1)=0 \end{gathered}$$
$\Rightarrow (\cos x-2)=0$ or, $(2\cos x-1)=0$
$\cos x=2$ is not possible.

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

(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) $4{\sin }^{2}x-8\cos x+1=0$
$$\begin{gathered} \Rightarrow 4-4{\cos }^{2}x-8\cos x+1=0 \\ \Rightarrow 4{\cos }^{2}x+8\cos x-5=0 \\ \Rightarrow 4{\cos }^{2}x+10\cos x-2\cos x-5=0 \\ \Rightarrow 2\cos x(2\cos x+5)-1(2\cos x+5)=0 \\ \Rightarrow (2\cos x-1)(2\cos x+5)=0 \end{gathered}$$
$\Rightarrow (2\mathrm{co}sx-1)=0$ or $(2\cos x+5)=0$
Now,
$2\cos x+5=0\Rightarrow \cos x=-\frac{5}{2}$ (It is not possible.)
$$\begin{gathered} \therefore 2\cos x-1=0 \\ \Rightarrow \cos x=\frac{1}{2} \\ \Rightarrow \cos x=\cos \frac{\mathrm{\pi }}{3} \\ \Rightarrow x=2n\pi \pm \frac{\mathrm{\pi }}{3},n\in Z \end{gathered}$$

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

(vii)
$$\begin{gathered} \cos 4x=\cos 2x \\ \Rightarrow 4x=2n\pi \pm 2x,n\in Z \end{gathered}$$
On taking positive sign, we have:
$$\begin{gathered} 4x=2n\pi +2x \\ \Rightarrow 2x=2n\pi \\ \Rightarrow x=n\pi ,n\in Z \end{gathered}$$
On taking negative sign, we have:
$$\begin{gathered} 4x=2n\pi -2x \\ \Rightarrow 6x=2n\pi \\ \Rightarrow x=\frac{n\mathrm{\pi }}{3},n\in Z \end{gathered}$$