Question

The number of solutions of the equation $ tan\, x + sec \,x = 2 cos \,x $ lying in the interval $ [0,2\pi] $ is

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

Answer: (D)

Given, $tan \,x + sec\, x = 2cos\,x$
$\frac{\left(1+sin\,x\right)}{cos\,x}=2\,cos\,x $
$1+sin\,x=2 cos^{2}\,x$
$1+sin\,x=2\left(1-sin^{2}\,x\right)$
$1+sin\,x=2-2\,sin^{2}\,x $
$2sin^{2}\,x+sin\,x-1=0$
$2sin^{2}\,x+2 sin\,x-sin\,x-1=0$
$2sin\,x\left(sin x+1\right)-1\left(sin\,x+1\right)=0$
$\left(sin\,x+1\right)\left(2\,sin\,x-1\right)=0
sin\,x=-1, 1 2$
$sin\,x=sin\,270^{\circ}, sin\,30^{\circ}, sin \,150^{\circ}$
$\Rightarrow x=\pi/ 6, 5\pi/ 6, 3\pi/ 2$
Hence there are three solutions of given equation in $\left[0, 2\pi\right]$