If x cos 5 x-1-0- where 0-x -2- find the value of

Given sec x cos 5x+1=0 ⇒1/cos x.cos 5x+1=0 ⇒ cos 5x+cos x=0 ⇒ 2cos 3x cos 2x=0 ⇒ cos 3x=0 or cos 2x=0 ⇒ 3x=(2n+1)π/2 or 2x=(2n+1)π/2 ⇒ x=(2n+1)π/6 or x=(2n+1)π/ ...

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Byju's Answer

Standard XII

Mathematics

Equations with Solutions at Boundary Values

If x cos 5 x+...

Question

$I f s e c x c o s 5 x + 1 = 0, w h e r e 0 < x \leq \frac{π}{2}, find the value of$

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Solution

$G i v e n s e c x c o s 5 x + 1 = 0 \Rightarrow \frac{1}{c o s x} . c o s 5 x + 1 = 0 \Rightarrow c o s 5 x + c o s x = 0 \Rightarrow 2 c o s 3 x c o s 2 x = 0 \Rightarrow c o s 3 x = 0 o r c o s 2 x = 0 \Rightarrow 3 x = (2 n + 1) \frac{π}{2} o r 2 x = (2 n + 1) \frac{π}{2} \Rightarrow x = (2 n + 1) \frac{π}{6} o r x = (2 n + 1) \frac{π}{4} \Rightarrow x = \frac{π}{6}, \frac{π}{2}, \frac{5 π}{6}, . . . . o r x = \frac{π}{4}, \frac{3 π}{4}, \frac{5 π}{4} . . . b u t 0 < x \leq \frac{π}{2} \Rightarrow x = \frac{π}{6} o r x = \frac{π}{4} o r x \frac{π}{2} H e n c e, t h e v a l u e s o f x a r e \frac{π}{6}, \frac{π}{4} .$

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If sec x cos 5x+1=0,where 0<x≤π2,find the value of

Given sec x cos 5x+1=0⇒1cos x.cos 5x+1=0⇒cos 5x+cos x=0⇒2cos 3x cos 2x=0⇒cos 3x=0 or cos 2x=0⇒3x=(2n+1)π2 or 2x=(2n+1)π2⇒x=(2n+1)π6 or x=(2n+1)π4⇒x=π6,π2,5π6,.... or x=π4,3π4,5π4...but 0<x≤ π2⇒x=π6 or x=π4 or x π2Hence,the values of x are π6,π4.

$I f s e c x c o s 5 x + 1 = 0, w h e r e 0 < x \leq \frac{π}{2}, find the value of$