The given equation is $$\tan x + \sec x = 2\cos x;$$
$$\eqalign{
& \Rightarrow \,\,\sin x + 1 = 2{\cos ^2}x \cr
& \Rightarrow \,\,\sin x + 1 = 2\left( {1 - {{\sin }^2}x} \right); \cr
& \Rightarrow \,\,2{\sin ^2}x + \sin x - 1 = 0; \cr
& \Rightarrow \,\,\left( {2\sin x - 1} \right)\left( {\sin x + 1} \right) = 0 \cr
& \Rightarrow \,\,\sin x = \frac{1}{2}, - 1.; \cr
& \Rightarrow \,\,x = {30^ \circ },{150^ \circ },{270^ \circ }. \cr} $$
2.
Let $$S = \left\{ {x \in \left( { - \pi ,\pi } \right):x \ne 0, \pm \frac{\pi }{2}} \right\}.$$ The sum of all distinct solutions of the equation $$\sqrt 3 \sec x + {\text{cosec}}\,x + 2\left( {\tan x - \cot x} \right) = 0$$ in the set $$S$$ is equal to
3.
The number of distinct solutions of $$\sin 5\theta \cdot \cos 3\theta = \sin 9\theta \cdot \cos 7\theta $$ in $$\left[ {0,\frac{\pi }{2}} \right]$$ is
The given equation can be written as
$$\eqalign{
& {e^{\sin x}} = 4 + \frac{1}{{{e^{\sin x}}}}\,\,\,.....\left( 1 \right) \cr
& {\text{Now, }} - 1 \leqslant \sin x \leqslant 1\,{\text{and }}e < 3 \cr
& \Rightarrow {e^{\sin x}} < 3 \cr
& \Rightarrow {\text{Again as we always have }}\frac{1}{{{e^{\sin x}}}} > 0 \cr
& \therefore 4 + \frac{1}{{{e^{\sin x}}}} > 4 \cr} $$
Thus the L.H.S. of $$\left( 1 \right) < 3$$ and R.H.S. of $$\left( 1 \right) > 4.$$
Hence there is no real values of $$x$$ which satisfy $$\left( 1 \right).$$
It follows that the given equation has no real solution.
10.
The number of solutions of the equation $$\tan x + \sec x = 2\cos x$$ lying in the interval $$\left[ {0,2\pi } \right]$$ is
Here, $$2{\sin ^2}x + \sin x - 1 = 0$$
$$ \Rightarrow \,\,\sin x = \frac{1}{2}, - 1.$$
But $$\sin x = - 1$$
$$ \Rightarrow \,\,x = \frac{{3\pi }}{2}$$
which does not satisfy the equation.