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Trigonometrical Equations
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The number of solution of the equation,$\sum\limits_{r = 1}^5 {\cos (r\,x)} $ $= 0$ lying in $(0, \pi)$ is :
A
$2$
B
$3$
C
$5$
D
more than $5$
Solution
$cos \,x + cos\, 2x + cos \,3x + cos \,4x + cos\, 5 x = 0$
$2\, cos \,3x \,cos \,2x + 2\, cos \,3x \,cos\, x + cos \,3x = 0$
$cos\, 3x [2 \,cos \,2x + 2 \,cos\, x – 1] = 0$
$\left. \begin{array}{l}x = (2n – 1)\frac{\pi }{6} \Rightarrow \frac{\pi }{6},\frac{{3\pi }}{6},\frac{{5\pi }}{6} = 3\\{2^{nd}} \,\, equation\,\, gives \,\,cos{\mkern 1mu} x = \frac{{1 \pm \sqrt 2 }}{4} = 2\end{array} \right] = 5$
Standard 11
Mathematics
