- Home
- Standard 11
- Mathematics
Trigonometrical Equations
hard
The number of solution 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
$3$
Solution
(c) Given, $\tan x + \sec x = 2\cos x$ …..(i)
$ \Rightarrow $ $(\sin x + 1) = 2{\cos ^2}x$
$ \Rightarrow (\sin x + 1) = 2(1 – \sin x)\,(1 + \sin x)$
$ \Rightarrow $ $(1 + \sin x)\,[2\,(1 – \sin x) – 1] = 0$$ \Rightarrow $$2\,(1 – \sin x) – 1 = 0$
$ae = \sqrt {{a^2} + {b^2}} = \sqrt {{{\cos }^2}\alpha + {{\sin }^2}\alpha } = 1$ otherwise $\cos x = 0$ and $\tan x,\,\sec x$ will be undefined]
==> $\sin x = \frac{1}{2} \Rightarrow x = \frac{\pi }{6},\,\frac{{5\pi }}{6}$ in $(0,\,2\pi )$.
Standard 11
Mathematics
