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

Hint: 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:
\[
  \sin {{\theta }} = 0 \Rightarrow {{\theta }} = n\pi \\
  \cos {{\theta }} = 0 \Rightarrow {{\theta }} = (2n + 1)\dfrac{\pi }{2} \\
  \tan {{\theta }} = 0 \Rightarrow {{\theta }} = n\pi \\
  \]