
The number of solutions of \[\sec x\cos 5x + 1 = 0\] in the interval $[0,2\pi ]$ is
A) 5
B) 8
C) 10
D) 12
Answer
571.2k+ views
Hint:
For this question we shall first convert the L.H.S. of the given equation as the sum of Cosine trigonometric ratios. Then use the trigonometric identity $\cos A + \cos B = 2\cos (\dfrac{{A + B}}{2})\cos (\dfrac{{A - B}}{2})$ to covert LHS as the product of Cosines. Then we shall equate each Cosine entity with zero & find all the values of $x$ which belong to the interval $[0,2\pi ]$ .
\[\sec x\cos 5x + 1 = 0\] … (1)
Complete step by step solution:
We already know that $\sec x = \dfrac{1}{{\cos x}}$, So in order to convert L.H.S. as the sum of Cosines divide the equation (1) by $\sec x$
$
\cos 5x + \dfrac{1}{{\sec x}} = 0 \\
\Rightarrow \cos 5x + \cos x = 0 \\
$
Now in order to convert them as product of Cosines we will be using $\cos A + \cos B = 2\cos (\dfrac{{A + B}}{2})\cos (\dfrac{{A - B}}{2})$
$
\Rightarrow 2\cos (\dfrac{{5x + x}}{2})\cos (\dfrac{{5x - x}}{2}) = 0 \\
\Rightarrow 2\cos (\dfrac{{6x}}{2})\cos (\dfrac{{4x}}{2}) = 0 \\
\Rightarrow 2\cos 3x\cos 2x = 0 \\
$
$ \Rightarrow \cos 3x = 0$ or $\cos x = 0$ … (2)
Now for $\cos 3x = 0$ we must find all the values of $x$ which lie in $[0,2\pi ]$. Then for $3x$ we must find all the values where Cosine is $0$ in $[0,6\pi ]$ .
$
3x = \dfrac{\pi }{2},\dfrac{{3\pi }}{2},\dfrac{{5\pi }}{2},\dfrac{{7\pi }}{2},\dfrac{{9\pi }}{2},\dfrac{{11\pi }}{2} \\
\Rightarrow x = \dfrac{\pi }{6},\dfrac{\pi }{2},\dfrac{{5\pi }}{6},\dfrac{{7\pi }}{6},\dfrac{{3\pi }}{2},\dfrac{{11\pi }}{6} \\
$
Now for $\cos 2x = 0$ we must find all the values of $x$ which lie in $[0,2\pi ]$. Then for $2x$ we must find all the values where Cosine is $0$ in $[0,4\pi ]$ .
$
2x = \dfrac{\pi }{2},\dfrac{{3\pi }}{2},\dfrac{{5\pi }}{2},\dfrac{{7\pi }}{2} \\
\Rightarrow x = \dfrac{\pi }{4},\dfrac{{3\pi }}{4},\dfrac{{5\pi }}{4},\dfrac{{7\pi }}{4} \\
\ $
Hence the equation (2) becomes
$(2) \Rightarrow x = \dfrac{\pi }{6},\dfrac{\pi }{4},\dfrac{{3\pi }}{4},\dfrac{{5\pi }}{4},\dfrac{{7\pi }}{4},\dfrac{\pi }{2},\dfrac{{5\pi }}{6},\dfrac{{7\pi }}{6},\dfrac{{3\pi }}{2},\dfrac{{11\pi }}{6}$
We can clearly see that all the values of $x$ lie in the interval $[0,2\pi ]$, Hence each value satisfies our answer.
The total no. of Solutions are the total values of $x$ which satisfy the conditions so there are 10 solutions.
Hence, the correct option is C.
Note:
While solving such questions, always try to convert each entity of the equation in Sine/Cosine functions. We already have established identities of Sine & Cosine functions sum/difference which allows the question to be solved easily.
For this question we shall first convert the L.H.S. of the given equation as the sum of Cosine trigonometric ratios. Then use the trigonometric identity $\cos A + \cos B = 2\cos (\dfrac{{A + B}}{2})\cos (\dfrac{{A - B}}{2})$ to covert LHS as the product of Cosines. Then we shall equate each Cosine entity with zero & find all the values of $x$ which belong to the interval $[0,2\pi ]$ .
\[\sec x\cos 5x + 1 = 0\] … (1)
Complete step by step solution:
We already know that $\sec x = \dfrac{1}{{\cos x}}$, So in order to convert L.H.S. as the sum of Cosines divide the equation (1) by $\sec x$
$
\cos 5x + \dfrac{1}{{\sec x}} = 0 \\
\Rightarrow \cos 5x + \cos x = 0 \\
$
Now in order to convert them as product of Cosines we will be using $\cos A + \cos B = 2\cos (\dfrac{{A + B}}{2})\cos (\dfrac{{A - B}}{2})$
$
\Rightarrow 2\cos (\dfrac{{5x + x}}{2})\cos (\dfrac{{5x - x}}{2}) = 0 \\
\Rightarrow 2\cos (\dfrac{{6x}}{2})\cos (\dfrac{{4x}}{2}) = 0 \\
\Rightarrow 2\cos 3x\cos 2x = 0 \\
$
$ \Rightarrow \cos 3x = 0$ or $\cos x = 0$ … (2)
Now for $\cos 3x = 0$ we must find all the values of $x$ which lie in $[0,2\pi ]$. Then for $3x$ we must find all the values where Cosine is $0$ in $[0,6\pi ]$ .
$
3x = \dfrac{\pi }{2},\dfrac{{3\pi }}{2},\dfrac{{5\pi }}{2},\dfrac{{7\pi }}{2},\dfrac{{9\pi }}{2},\dfrac{{11\pi }}{2} \\
\Rightarrow x = \dfrac{\pi }{6},\dfrac{\pi }{2},\dfrac{{5\pi }}{6},\dfrac{{7\pi }}{6},\dfrac{{3\pi }}{2},\dfrac{{11\pi }}{6} \\
$
Now for $\cos 2x = 0$ we must find all the values of $x$ which lie in $[0,2\pi ]$. Then for $2x$ we must find all the values where Cosine is $0$ in $[0,4\pi ]$ .
$
2x = \dfrac{\pi }{2},\dfrac{{3\pi }}{2},\dfrac{{5\pi }}{2},\dfrac{{7\pi }}{2} \\
\Rightarrow x = \dfrac{\pi }{4},\dfrac{{3\pi }}{4},\dfrac{{5\pi }}{4},\dfrac{{7\pi }}{4} \\
\ $
Hence the equation (2) becomes
$(2) \Rightarrow x = \dfrac{\pi }{6},\dfrac{\pi }{4},\dfrac{{3\pi }}{4},\dfrac{{5\pi }}{4},\dfrac{{7\pi }}{4},\dfrac{\pi }{2},\dfrac{{5\pi }}{6},\dfrac{{7\pi }}{6},\dfrac{{3\pi }}{2},\dfrac{{11\pi }}{6}$
We can clearly see that all the values of $x$ lie in the interval $[0,2\pi ]$, Hence each value satisfies our answer.
The total no. of Solutions are the total values of $x$ which satisfy the conditions so there are 10 solutions.
Hence, the correct option is C.
Note:
While solving such questions, always try to convert each entity of the equation in Sine/Cosine functions. We already have established identities of Sine & Cosine functions sum/difference which allows the question to be solved easily.
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