Question

If $(\cos x - \sin x)\left( 2\tan x + \frac{1}{\cos x} \right) + 2 = 0$ then $x =$

• A. $2n\pi \pm \frac{\pi}{3}$
• B. $n\pi \pm \frac{\pi}{3}$
• C. $2n\pi \pm \frac{\pi}{6}$
• D. None of these

Answer 1

Answer: (A)

$2n\pi \pm \frac{\pi}{3}$

Answer 2

Let $t = \tan\frac{x}{2},$ and using the formula. We get,

$\left\{ \frac { 1 - \tan ^ { 2 } \frac { x } { 2 } } { 1 + \tan ^ { 2 } \frac { x } { 2 } } - \frac { 2 \tan \frac { x } { 2 } } { 1 + \tan ^ { 2 } \frac { x } { 2 } } \right\}$ $\left\{ \frac{\mathbf{4}\mathbf{\tan}\frac{\mathbf{x}}{\mathbf{2}}}{\mathbf{1 -}\mathbf{\tan}^{\mathbf{2}}\frac{\mathbf{x}}{\mathbf{2}}}\mathbf{+}\frac{\mathbf{1 +}\mathbf{\tan}^{\mathbf{2}}\frac{\mathbf{x}}{\mathbf{2}}}{\mathbf{1 -}\mathbf{\tan}^{\mathbf{2}}\frac{\mathbf{x}}{\mathbf{2}}} \right\}\mathbf{+ 2 = 0}$ $\left( \frac { 1 - t ^ { 2 } } { 1 + t ^ { 2 } } - \frac { 2 t } { 1 + t ^ { 2 } } \right)$ $\left( \frac{\mathbf{4t}}{\mathbf{1}\mathbf{-}\mathbf{t}^{\mathbf{2}}}\mathbf{+}\frac{\mathbf{1 +}\mathbf{t}^{\mathbf{2}}}{\mathbf{1}\mathbf{-}\mathbf{t}^{\mathbf{2}}} \right)\mathbf{+ 2 = 0}$

$\mathbf{\Rightarrow}$ $\frac{\mathbf{3}\mathbf{t}^{\mathbf{4}}\mathbf{+ 6}\mathbf{t}^{\mathbf{3}}\mathbf{+ 8}\mathbf{t}^{\mathbf{2}}\mathbf{-}\mathbf{2t}\mathbf{-}\mathbf{3}}{\mathbf{(}\mathbf{t}^{\mathbf{2}}\mathbf{+ 1)(1}\mathbf{-}\mathbf{t}^{\mathbf{2}}\mathbf{)}}\mathbf{= 0}$

Its roots are; $t_{1} = \frac{1}{\sqrt{3}}$and $t_{2} = - \frac{1}{\sqrt{3}}.$

Thus the solution of the equation reduces to that of two elementary equations,

$\tan\frac{x}{2} = \frac{1}{\sqrt{3}},\tan\frac{x}{2} = - \frac{1}{\sqrt{3}}$ $\Rightarrow$ $\frac{x}{2} = n\pi \pm \frac{\pi}{6}$

$\Rightarrow x = 2n\pi \pm \frac{\pi}{3},$ is required solution.