Question

If \[\sec x\cos 5x + 1 = 0\], where \[0 < x < 2\pi \], then\[{\rm{x}} = \]
A.\[\frac{\pi }{5},\frac{\pi }{5}\]
B. \[\frac{\pi }{5}\]
C. \[\frac{\pi }{4}\]
D. None of these

Answer 1

Hint
Equation's roots: \[a{x^n} + b{x^{n - 1}} + .........c = 0\] are those x-values that, when substituted into the equation, result in the equation's value being equal to zero. Write \[\left( {xy} \right) = 0\] if y is a root of the equation is an ingredient in the equation. We put\[x = y\]on the equation for the factor is one of the equation's roots. In this case, at first we can write the given equation as \[\cos 5x + \cos x = 0\]
After that, apply the below formula to proceed with determining \[\cos A + \cos B = 2\cos \left( {\frac{{A + B}}{2}} \right)\cos \left( {\frac{{A - B}}{2}} \right)\]
Now, we get\[2 \cdot \cos 3x \cdot \cos 2x = 0\]. After that we solve the equation, for the given interval.
Formula used:
[\cos A + \cos B = 2\cos \left( {\frac{{A + B}}{2}} \right)\cos \left( {\frac{{A - B}}{2}} \right)\]
Complete step-by-step Solution:
First, we formulate the given equation:
\[\sec x\cos 5x + 1 = 0\]
Subtract either side of the equation by\[{\rm{1}}\]:
\[ \Rightarrow \sec x\cos 5x = - 1\]
Divide either side of the equation by\[\sec \left( x \right)\]:
Use the basic trigonometry identity\[\frac{1}{{\sec \left( x \right)}} = \cos \left( x \right)\]:
\[ = - \cos \left( x \right)\]
After applying the trigonometry identity, equation becomes
\[ \Rightarrow \cos 5x = - \cos x\]
Subtract\[ - \cos \left( x \right)\]from both sides:
\[\cos \left( {5x} \right) + \cos \left( x \right) = 0\]
Solving each part separately:
\[\cos \left( {2x} \right) = 0\]or\[\cos \left( {3x} \right) = 0\]
Rewrite using trigonometry identities:
\[2\cos \left( {2x} \right)\cos \left( {3x} \right) = 0\]
Derive the solutions:
\[ \Rightarrow x = \frac{{(2n + 1)\pi }}{6}{\rm{ or }}\frac{{(2n - 1)\pi }}{4}\]
Where, “n” is any integer.
The general solution obtained is,
\[ \Rightarrow 5x = 2n\pi \pm (\pi - x)\]
This can also be written as
\[ \Rightarrow x = \frac{{(2n + 1)\pi }}{6}{\rm{ or }}\frac{{(2n - 1)\pi }}{4}\]
On substituting the values of n we get,
From these x values in the range\[\left[ {0,{\rm{ }}2} \right]\], one can get
\[x = \frac{\pi }{4},\frac{\pi }{2},\frac{{3\pi }}{4},\frac{{5\pi }}{6},\frac{{5\pi }}{4},\frac{{7\pi }}{6},\frac{{7\pi }}{4},\frac{{9\pi }}{6},\frac{{11\pi }}{6}\]
Therefore, for\[\sec x\cos 5x + 1 = 0\], the value of\[{\rm{x}} = \frac{\pi }{4}\]
Hence, the option C is correct.

Note
One should be familiar with the fundamental trigonometric formulas as well as the universal formula for all trigonometric ratios. As they don't check for the beginning condition, students are likely to make the error of writing that the equation has three roots. Nonetheless, those are the\[sin(x)\]function's values, rresulting from the quadratic equation being solved.