Question

The number of solution of $\sec x \cdot \cos 5x + 1 = 0$ in the interval $\left[ {0,2\pi } \right]$ is
A.5
B.8
C.10
D.12

Answer 1

At first, we can write the given equation as $\cos 5x + \cos x = 0$
Then we have to apply the formula
$\cos A + \cos B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)$
We get $2 \cdot \cos 3x \cdot \cos 2x = 0$
After that we solve the equation, for the given interval.

Complete step-by-step answer:
Firstly we write the given equation, which is $\sec x \cdot \cos 5x + 1 = 0$
This equation can also be written as $\cos 5x + \cos x = 0$; since $\cos x = \dfrac{1}{{\sec x}} \Rightarrow \sec x = \dfrac{1}{{\cos x}}$.
Now, we apply the formula,
$\cos A + \cos B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)$
Then we obtain from the last equation

$$2\cos \left( {\dfrac{{5 + 1}}{2}} \right)x \cdot \cos \left( {\dfrac{{5 - 1}}{2}} \right)x = 0 \\\ \Rightarrow \cos \dfrac{{6x}}{2} \cdot \cos \dfrac{{4x}}{2} = 0\left[ {\because 2 \ne 0} \right] \\\ \Rightarrow \cos 3x \cdot \cos 2x = 0 \\\$$

This implies that
$\cos 3x = 0...........(2)$
And $\cos 2x = 0...........(3)$
Again, we know the formula,

$$\cos x = 0 \\\ \Rightarrow x = \left( {2n + 1} \right)\dfrac{\pi }{2} \\\$$

Where, n is any integer.
Therefore, from the equation (2) we get

$$3x = \left( {2n + 1} \right)\dfrac{\pi }{2} \\\ \Rightarrow x = \left( {2n + 1} \right)\dfrac{\pi }{6};where,n = 0, \pm 1, \pm 2,..... \\\$$

On substituting the values of n we get,
i.e. x = \dfrac{\pi }{6}, - \dfrac{\pi }{6},\dfrac{\pi }{2}, - \dfrac{\pi }{2},\dfrac{{5\pi }}{6}, - \dfrac{{5\pi }}{6},....$$$$ = \pm \dfrac{\pi }{6}, \pm \dfrac{\pi }{2}, \pm \dfrac{{5\pi }}{6},....
Also we get from the equation (3)

$$2x = \left( {2n + 1} \right)\dfrac{\pi }{2} \\\ \Rightarrow x = \left( {2n + 1} \right)\dfrac{\pi }{4};where,n = 0, \pm 1, \pm 2,..... \\\$$

On substituting the values of n we get,
i.e. x = \dfrac{\pi }{4}, - \dfrac{\pi }{4},\dfrac{{3\pi }}{4}, - \dfrac{{3\pi }}{4},\dfrac{{5\pi }}{4}, - \dfrac{{5\pi }}{4},....$$$$ = \pm \dfrac{\pi }{4}, \pm \dfrac{\pi }{4}, \pm \dfrac{{5\pi }}{4},....
From these values of x which are in the interval $\left[ {0,2\pi } \right]$ is $x = \dfrac{\pi }{6},\dfrac{\pi }{4},\dfrac{\pi }{2},\dfrac{{3\pi }}{4},\dfrac{{3\pi }}{6},\dfrac{{5\pi }}{6},\dfrac{{5\pi }}{4},\dfrac{{7\pi }}{4},\dfrac{{7\pi }}{6},\dfrac{{9\pi }}{6}$
Hence option C is the answer.

Note: One should know the basic formulas of trigonometry, and also know the general formula of all the trigonometric ratios.
They are:

$$\sin {{\theta }} = 0 \Rightarrow {{\theta }} = n\pi \\\ \cos {{\theta }} = 0 \Rightarrow {{\theta }} = (2n + 1)\dfrac{\pi }{2} \\\ \tan {{\theta }} = 0 \Rightarrow {{\theta }} = n\pi \\\$$