L e t f ( x ) = sin 23 x − cos 22 x a n d g ( x ) = 1 + 1 2 tan − 1 ∣ x ∣ , t h e n o . o f v a l u e s o f x i n i n t e r v a l [ − 10 π , 20 π ] s a t i s f y i n g t h e e q u a t i o n f ( x ) = s g n ( g ( x ) ) i s Let\quad f\left( x \right) =\sin ^{ 23 }{ x-\cos ^{ 22 }{ x } } \quad and\quad g\left( x \right) =1+\frac { 1 }{ 2 } \tan ^{ -1 }{ \left| x \right| } ,\\ the\quad no.\quad of\quad values\quad of\quad x\quad in\quad interval\quad \left[ -10\pi ,20\pi \right] \quad satisfying\\ the\quad equation\quad f\left( x \right) =sgn\left( g\left( x \right) \right) \quad is L e t f ( x ) = sin 2 3 x − cos 2 2 x a n d g ( x ) = 1 + 2 1 tan − 1 ∣ x ∣ , t h e n o . o f v a l u e s o f x i n i n t e r v a l [ − 1 0 π , 2 0 π ] s a t i s f y i n g t h e e q u a t i o n f ( x ) = s g n ( g ( x ) ) i s
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Let's observe g ( x ) g(x) g ( x ) . g ( x ) = 1 + 1 2 tan − 1 ∣ x ∣ g(x) = 1+\frac{1}{2} \tan ^{-1}|x| g ( x ) = 1 + 2 1 tan − 1 ∣ x ∣ Whatever be the value of x x x , g ( x ) g(x) g ( x ) is always + + + ve. Hence s g n ( g ( x ) ) sgn(g(x)) s g n ( g ( x ) ) is always 1 1 1 . Hence our task is reduced to find the solution of f ( x ) = 1 f(x) = 1 f ( x ) = 1 ⟹ sin 23 x − cos 22 x = 1 \implies \sin^{23}x - \cos^{22} x = 1 ⟹ sin 2 3 x − cos 2 2 x = 1 ⟹ sin 23 x = 1 + cos 22 x \implies \sin^{23} x = 1+ \cos^{22}x ⟹ sin 2 3 x = 1 + cos 2 2 x Now we observe that R H S ≥ 1 RHS \geq 1 R H S ≥ 1 , but L H S ≤ 1 LHS \leq 1 L H S ≤ 1 . Hence only solution is that 1 + cos 22 x = 1 a n d sin 23 x = 1 1+\cos^{22}x= 1 and \sin^{23} x =1 1 + cos 2 2 x = 1 a n d sin 2 3 x = 1 Solving above we get solution of the form x = 2 n π + π 2 x = 2n\pi + \dfrac{\pi}{2} x = 2 n π + 2 π The values of n n n which satisfy the given domain are { − 5 , − 4 , − 3 , … 7 , 8 , 9 } \{-5,-4,-3, \ldots 7,8,9\} { − 5 , − 4 , − 3 , … 7 , 8 , 9 } .
Nice solution @Prakhar Gupta
sgn(g(X))=1 as g(X)>=1 so f(X)=1 then x=2n pi + pi/2
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Let's observe g ( x ) . g ( x ) = 1 + 2 1 tan − 1 ∣ x ∣ Whatever be the value of x , g ( x ) is always + ve. Hence s g n ( g ( x ) ) is always 1 . Hence our task is reduced to find the solution of f ( x ) = 1 ⟹ sin 2 3 x − cos 2 2 x = 1 ⟹ sin 2 3 x = 1 + cos 2 2 x Now we observe that R H S ≥ 1 , but L H S ≤ 1 . Hence only solution is that 1 + cos 2 2 x = 1 a n d sin 2 3 x = 1 Solving above we get solution of the form x = 2 n π + 2 π The values of n which satisfy the given domain are { − 5 , − 4 , − 3 , … 7 , 8 , 9 } .